Class 12 Electrostatics Masterclass

Electric Field Concept

Electric field tells us how a charge changes the space around it. At any point, it gives both the strength and direction of force that a small positive test charge would experience.

  • CBSE Class 12
  • NEET
  • JEE Main
  • JEE Advanced
  • IB Physics
  • ICSE
  • IGCSE
  • British Curriculum

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Electric field around a positive source charge + Direction of E = force on positive test charge

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Concept 1

Electric Field Definition

Before writing a formula, remember the experiment behind it: place a very small positive test charge at a point and measure the electric force on it. Electric field is the force per coulomb at that point.

Formula with meaning

If a positive test charge q0 experiences force F at a point, the electric field intensity E is:

E = F / q0
  • E is a vector quantity.
  • SI unit: N/C. It is also equal to V/m.
  • Direction of E is the direction of force on a positive test charge.
  • The test charge should be small so it does not disturb the source charge arrangement.
Exam tip: If the charge placed in the field is negative, its force is opposite to E, but E itself does not reverse.

Physical picture

A source charge creates electric field in the surrounding space. Another charge does not need to touch the source charge; it responds to the field already present at its location.

This idea is central to electrostatics because it turns charge-force problems into local field problems. Once E is known, force on any charge q is found from F = qE.

F = qE

Concept 2

Electric Field Due to a Point Charge

The field due to a point charge comes from Coulomb's law. Divide the force on a small positive test charge by that test charge, and q0 cancels out.

Derivation before formula

Let source charge Q be fixed at a point. Place a small positive test charge q0 at distance r.
Coulomb force on q0 is
F = (1 / 4πε0) x Qq0 / r^2
Electric field is force per unit positive test charge:
E = F / q0
After cancellation of q0:
E = (1 / 4πε0) x Q / r^2 = kQ/r^2

Direction and variation

  • For +Q, electric field is radially outward.
  • For -Q, electric field is radially inward.
  • The magnitude depends on |Q| and r, while direction depends on sign and position.
  • Doubling distance makes field one-fourth; tripling distance makes it one-ninth.
E ∝ Q and E ∝ 1/r^2
Exam tip: In MCQs, signs are best handled by a direction diagram. Use formula for magnitude and arrows for direction.

Concept 3

Electric Field Lines and Their Properties

Field lines are imaginary curves used to visualize electric field. The tangent gives direction, while the closeness of lines shows relative strength.

Positive source charge

Outward field lines for positive charge +

Negative source charge

Inward field lines for negative charge -

Dipole field lines

Field lines from positive to negative charge +-

Conductor surface

Electric field lines perpendicular to conductor surface metal

Properties of electric field lines

  • They start from positive charge and end on negative charge or infinity.
  • Tangent at any point gives the direction of electric field.
  • They never intersect because field direction at a point is unique.
  • They do not form closed loops in electrostatics.
  • Closer spacing means stronger electric field.

Exam tips for field lines

  • For +Q, draw arrows outward; for -Q, draw arrows inward.
  • For equal and opposite charges, lines connect from + to -.
  • For equal like charges, lines bend away from the region between them.
  • At a conductor surface, field lines must be normal to the surface.

Concept 4

Superposition of Electric Field

The net electric field due to many charges is the vector sum of the electric fields produced by individual charges at the same point.

Vector addition

Calculate E1, E2, E3 separately and add them with directions. Equal magnitudes do not always mean cancellation unless directions are opposite.

Symmetry shortcut

At the centre of a square, triangle, or hexagon with equal charges, symmetry often makes net field zero without long calculation.

Null point logic

For like charges, zero field can lie between them. For unequal unlike charges, it lies outside on the side of the smaller magnitude charge.

Concept 5

Electric Dipole and Dipole Moment

An electric dipole consists of two equal and opposite charges separated by a small distance. It is the first important example where vector superposition produces a new field pattern.

Dipole moment

If the charges are -q and +q separated by distance 2a, the dipole moment is:

p = q x 2a
  • Direction of p is from -q to +q.
  • SI unit is C m.
  • Dipoles are important in torque, dielectrics, molecular polarity, and electric potential.

Electric dipole diagram

Dipole moment from negative charge to positive charge -q +q dipole moment p 2a

Concept 6

Axial Field and Equatorial Field of a Dipole

Always decide where the observation point lies before using a formula. Axial points lie on the dipole axis; equatorial points lie on the perpendicular bisector.

Axial point

Place -q and +q separated by 2a, with centre O. Let P be on the axis at distance r from O.
Distances from the two charges are r + a and r - a.
The two fields act along the same line and subtract in magnitude.
E_axial = (1 / 4πε0) x [2pr / (r^2 - a^2)^2]
For r >> a: E_axial = (1 / 4πε0) x 2p/r^3

Equatorial point

Let P be on the perpendicular bisector at distance r from the centre.
Distances from +q and -q are equal: sqrt(r^2 + a^2).
Perpendicular components cancel; components along the dipole axis add opposite to p.
E_equatorial = (1 / 4πε0) x [p / (r^2 + a^2)^(3/2)]
For r >> a: E_equatorial = (1 / 4πε0) x p/r^3
ComparisonAxial pointEquatorial point
PositionOn dipole axisOn perpendicular bisector
Exact formulak x 2pr/(r^2 - a^2)^2k x p/(r^2 + a^2)^(3/2)
Short dipole formula2kp/r^3kp/r^3
Direction trapAlong axis; commonly along p on the outer axial sideOpposite to p
At same far distanceTwice the equatorial magnitudeHalf the axial magnitude

Concept 7

Moving Charge, Conductors and Common Mistakes

Electrostatics mostly studies charges at rest. Moving charges require extra care because magnetic effects also appear.

Electric field due to moving charge

A moving charge is still a source of electric field, but it is also associated with magnetic field. Detailed moving-charge field treatment belongs to electromagnetism and relativity, so school electrostatics first focuses on stationary point charges and dipoles.

Electric field inside conductor

Inside a conductor in electrostatic equilibrium, E = 0. At the surface, electric field is perpendicular; a tangential component would make free charges move.

Common language correction

Use “moving charge” or “charge in motion.” Avoid writing “current-carrying charge” as a standard phrase in electrostatics answers.

Source vs test charge

Source charges create E. A test charge only detects E. Do not use q0 as the source charge in point-charge formula.

Magnitude vs direction

Use E = k|Q|/r^2 for magnitude. Decide inward or outward using the sign of Q and a diagram.

Vector addition

Electric fields can cancel even when potential does not. Add field vectors, not just magnitudes.

Formula Summary

Important Electric Field Formulas

Use these formulas after understanding the direction and the geometry of the point where electric field is required.

E = F/q0Definition of electric field intensity.
F = qEForce on charge q placed in field E.
E = kQ/r^2Magnitude of field due to a point charge.
k = 1/(4πε0)Electrostatic constant in vacuum.
p = q x 2aElectric dipole moment.
E_axial = 2kp/r^3Short dipole, far axial point.
E_equatorial = kp/r^3Short dipole, far equatorial point.
E_inside conductor = 0Electrostatic equilibrium condition.

Original Practice Bank

Electric Field Questions for NEET, IIT Main, JEE Advanced, IB, ICSE and IGCSE

All important questions below are written directly in HTML for better readability and indexing. They are original exam-style questions, not copied previous-year claims.

25 Unique MCQs

NEET Electric Field Questions

NEET-level multiple-choice questions on E = F/q0, point charges, field direction, field lines, conductors, dipoles and force on charges.

NEET MCQ 1

NEET Electric Field Question 1: Force Per Unit Test Charge

A +2 microcoulomb test charge experiences a force of 6.0 x 10^-3 N at a point. The electric field at that point is:

  1. 1.2 x 10^-8 N/C
  2. 3.0 x 10^3 N/C
  3. 12 x 10^-3 N/C
  4. 8.0 x 10^3 N/C
Correct answer: B
Short explanation: Use E = F/q0. Thus E = (6.0 x 10^-3)/(2.0 x 10^-6) = 3.0 x 10^3 N/C.
NEET MCQ 2

NEET Electric Field Question 2: Direction Near a Positive Source Charge

A positive point charge is fixed at the origin. At a point on the positive x-axis, the electric field is directed:

  1. Along negative x-axis
  2. Along positive x-axis
  3. Along positive y-axis
  4. Perpendicular to the radius
Correct answer: B
Short explanation: The electric field due to a positive charge is radially outward, so at a point on +x it points along +x.
NEET MCQ 3

NEET Electric Field Question 3: Field Near a Negative Source Charge

A negative source charge is placed at the origin. At a point to its right, the electric field points:

  1. Away from the charge
  2. Toward the charge
  3. Vertically upward
  4. It has no definite direction
Correct answer: B
Short explanation: A negative charge attracts a positive test charge, so the field direction is inward, toward the source charge.
NEET MCQ 4

NEET Electric Field Question 4: Inverse Square Variation

If the distance from an isolated point charge is doubled, the magnitude of electric field becomes:

  1. E/2
  2. 2E
  3. E/4
  4. 4E
Correct answer: C
Short explanation: For a point charge, E is proportional to 1/r^2. Doubling r makes the field one-fourth.
NEET MCQ 5

NEET Electric Field Question 5: Field Line Origin and Termination

In electrostatics, electric field lines generally:

  1. Start on negative charges and end on positive charges
  2. Start on positive charges and end on negative charges
  3. Always form closed loops
  4. Always intersect near strong fields
Correct answer: B
Short explanation: Electrostatic field lines originate from positive charges and terminate on negative charges or infinity.
NEET MCQ 6

NEET Electric Field Question 6: Meaning of Crowded Field Lines

A region where electric field lines are drawn very close to one another represents:

  1. Weak electric field
  2. Strong electric field
  3. Zero electric potential only
  4. Magnetic field only
Correct answer: B
Short explanation: The density of field lines indicates relative strength of electric field.
NEET MCQ 7

NEET Electric Field Question 7: Field Inside an Electrostatic Conductor

The electric field inside the material of a conductor in electrostatic equilibrium is:

  1. Maximum
  2. Zero
  3. Equal to external field always
  4. Infinite
Correct answer: B
Short explanation: Free charges rearrange until the net electric field inside the conducting material becomes zero.
NEET MCQ 8

NEET Electric Field Question 8: Field Lines at a Conductor Surface

At the surface of a conductor in electrostatic equilibrium, electric field lines are:

  1. Parallel to the surface
  2. Perpendicular to the surface
  3. Circular around the surface
  4. Randomly oriented
Correct answer: B
Short explanation: A tangential electric field would move free charges, so only the normal component remains at equilibrium.
NEET MCQ 9

NEET Electric Field Question 9: Dipole Moment Definition

For an electric dipole with charges +q and -q separated by distance 2a, dipole moment is:

  1. q/a directed from +q to -q
  2. q x 2a directed from -q to +q
  3. 2a/q directed from -q to +q
  4. q x a directed perpendicular to the axis
Correct answer: B
Short explanation: Electric dipole moment p = q(2a), and its direction is conventionally from negative charge to positive charge.
NEET MCQ 10

NEET Electric Field Question 10: SI Unit of Dipole Moment

The SI unit of electric dipole moment is:

  1. N/C
  2. C m
  3. V/m
  4. C/m
Correct answer: B
Short explanation: Dipole moment equals charge multiplied by separation, so the unit is coulomb metre.
NEET MCQ 11

NEET Electric Field Question 11: Short Dipole Axial Formula

For a short electric dipole, the magnitude of electric field at an axial point far from the dipole is:

  1. kp/r^2
  2. 2kp/r^3
  3. kp/r^3
  4. 2kq/r^2
Correct answer: B
Short explanation: The far axial field of a short dipole is E_axial = 2kp/r^3.
NEET MCQ 12

NEET Electric Field Question 12: Equatorial Dipole Direction

At an equatorial point of a dipole, the net electric field is directed:

  1. Along dipole moment p
  2. Opposite to dipole moment p
  3. Always zero
  4. Perpendicular to the dipole axis
Correct answer: B
Short explanation: At the equatorial point, perpendicular components cancel and axial components add opposite to p.
NEET MCQ 13

NEET Electric Field Question 13: Axial and Equatorial Ratio

At the same large distance r from a short dipole, the ratio of axial field to equatorial field is:

  1. 1:1
  2. 1:2
  3. 2:1
  4. 4:1
Correct answer: C
Short explanation: E_axial = 2kp/r^3 and E_equatorial = kp/r^3, so the ratio is 2:1.
NEET MCQ 14

NEET Electric Field Question 14: Force on a Negative Charge

A uniform electric field of 200 N/C is directed east. The force on a -3 microcoulomb charge is:

  1. 6.0 x 10^-4 N east
  2. 6.0 x 10^-4 N west
  3. 600 N east
  4. 600 N west
Correct answer: B
Short explanation: Magnitude |F| = |q|E = 3 x 10^-6 x 200 = 6.0 x 10^-4 N; negative charge feels force opposite to E.
NEET MCQ 15

NEET Electric Field Question 15: Field Magnitude of a Point Charge

A +4 microcoulomb charge is observed from a distance of 2 m. Taking k = 9 x 10^9 SI units, the electric field magnitude is:

  1. 9.0 x 10^3 N/C
  2. 1.8 x 10^4 N/C
  3. 3.6 x 10^4 N/C
  4. 4.5 x 10^3 N/C
Correct answer: A
Short explanation: E = kQ/r^2 = (9 x 10^9)(4 x 10^-6)/4 = 9.0 x 10^3 N/C.
NEET MCQ 16

NEET Electric Field Question 16: Test Charge Independence

If the positive test charge used to measure electric field is doubled, the electric field due to fixed source charges:

  1. Doubles
  2. Becomes half
  3. Remains unchanged
  4. Becomes zero
Correct answer: C
Short explanation: The force on the test charge doubles, but E = F/q0 remains the same for the same source arrangement.
NEET MCQ 17

NEET Electric Field Question 17: Zero Field and Force

At a point where the net electric field is zero, the electric force on any small charge placed there is:

  1. Always zero
  2. Always infinite
  3. Equal to charge only
  4. Opposite to velocity
Correct answer: A
Short explanation: Electric force is F = qE. If E = 0, then F = 0 for any value of q.
NEET MCQ 18

NEET Electric Field Question 18: Non-Intersection of Field Lines

Electric field lines never intersect because at an intersection point:

  1. Electric field would be zero only
  2. Potential would become infinite
  3. Electric field would have two directions
  4. Charge would disappear
Correct answer: C
Short explanation: The tangent to a field line gives field direction; two crossing lines would imply two directions at one point.
NEET MCQ 19

NEET Electric Field Question 19: Moving Charge Field

A charge moving steadily with respect to an observer is associated with:

  1. Only electric field
  2. Only magnetic field
  3. Both electric and magnetic fields
  4. No field at all
Correct answer: C
Short explanation: At school level, electrostatics focuses on rest charges, but moving charges are associated with magnetic effects too.
NEET MCQ 20

NEET Electric Field Question 20: Superposition Principle

The net electric field at a point due to many charges is obtained by:

  1. Adding only magnitudes
  2. Multiplying all fields
  3. Vector addition of individual fields
  4. Taking the largest field only
Correct answer: C
Short explanation: Electric field is a vector, so direction must be included while adding contributions.
NEET MCQ 21

NEET Electric Field Question 21: Midpoint of Equal Like Charges

Two equal positive charges are placed symmetrically on a line. At the midpoint between them, the net electric field is:

  1. Toward left charge
  2. Toward right charge
  3. Zero
  4. Perpendicular to the line
Correct answer: C
Short explanation: The two fields at the midpoint are equal in magnitude and opposite in direction, so they cancel.
NEET MCQ 22

NEET Electric Field Question 22: Midpoint of Equal Unlike Charges

At the midpoint between equal charges +Q and -Q, the net electric field is:

  1. Zero
  2. From +Q toward -Q
  3. From -Q toward +Q
  4. Perpendicular to the line joining charges
Correct answer: B
Short explanation: At the midpoint, field due to +Q points away from +Q and field due to -Q points toward -Q; both point from + to -.
NEET MCQ 23

NEET Electric Field Question 23: Potential Zero but Field Non-Zero

At the midpoint of equal and opposite charges +Q and -Q, which statement is correct?

  1. Potential is zero and electric field is also zero
  2. Potential is zero but electric field is non-zero
  3. Potential is non-zero but electric field is zero
  4. Both quantities are infinite
Correct answer: B
Short explanation: Potential cancels as a scalar, but electric field vectors add in the direction from positive charge to negative charge.
NEET MCQ 24

NEET Electric Field Question 24: Shielding by a Conductor

Electrostatic shielding by a closed conductor is mainly explained by:

  1. Zero electric field inside the conductor material
  2. Closed-loop field lines
  3. Increase of field inside metal
  4. Conversion of charge into mass
Correct answer: A
Short explanation: Free charges redistribute on the conductor so that the electric field inside conducting material becomes zero.
NEET MCQ 25

NEET Electric Field Question 25: Force Direction on a Positive Charge

A +5 microcoulomb charge is placed in an electric field directed upward. The electric force on it is:

  1. Upward
  2. Downward
  3. Zero because charge is small
  4. Horizontal
Correct answer: A
Short explanation: For positive q, F = qE has the same direction as E.

25 Unique Questions

JEE Main / IIT Main Electric Field Questions

Numerical and conceptual problems using vector addition, point-charge variation, dipole variation, superposition, symmetry and conductor logic.

JEE Main / IIT Main 1

JEE Main Electric Field Question 1: Scaling with Charge and Distance

A charge Q produces field E at distance r. What field will charge 3Q produce at distance 2r?

Answer: 3E/4
Formula used: E = kQ/r^2
  1. Original field is E = kQ/r^2.
  2. New field is E' = k(3Q)/(2r)^2.
  3. Therefore E' = 3E/4.
JEE Main / IIT Main 2

JEE Main Electric Field Question 2: Perpendicular Vector Addition

Two electric fields at a point are 3 N/C east and 4 N/C north. Find the magnitude of the resultant field.

Answer: 5 N/C
Formula used: E_net = sqrt(E_x^2 + E_y^2) for perpendicular vectors
  1. The fields are perpendicular.
  2. E_net = sqrt(3^2 + 4^2).
  3. E_net = 5 N/C.
JEE Main / IIT Main 3

JEE Main Electric Field Question 3: Symmetric Like Charges

Two equal positive charges are placed at x = +a and x = -a. What is the electric field at the origin?

Answer: Zero
Formula used: Net electric field is vector sum of individual fields
  1. Each charge produces the same field magnitude at the origin.
  2. The directions are opposite along the x-axis.
  3. The two contributions cancel exactly.
JEE Main / IIT Main 4

JEE Main Electric Field Question 4: Field Between Unlike Charges

Charges +Q at x = -a and -Q at x = +a are placed on the x-axis. Determine the direction of field at the origin.

Answer: Along positive x-axis, from +Q toward -Q
Formula used: Field due to +Q is away from +Q; field due to -Q is toward -Q
  1. At the origin, field due to +Q points to the right.
  2. Field due to -Q also points to the right.
  3. So the net field is along +x.
JEE Main / IIT Main 5

JEE Main Electric Field Question 5: One Charge Removed from a Square

Four equal positive charges at the corners of a square give zero field at the centre. If one charge is removed, what is the net field due to the remaining three?

Answer: Equal in magnitude and opposite in direction to the field that would have been produced by the removed charge alone
Formula used: E_total(all four) = 0
  1. Let the field due to the removed charge be E_r.
  2. For all four charges, E_r + E_remaining = 0.
  3. Hence E_remaining = -E_r.
JEE Main / IIT Main 6

JEE Main Electric Field Question 6: Centre of an Equilateral Triangle

Three equal positive charges are placed at the vertices of an equilateral triangle. Find the electric field at its centroid.

Answer: Zero
Formula used: Fields of equal magnitude separated by 120 degrees cancel
  1. The centroid is equidistant from all three charges.
  2. The three field vectors have equal magnitude and symmetric directions.
  3. Their vector sum is zero.
JEE Main / IIT Main 7

JEE Main Electric Field Question 7: Null Point Between Unequal Like Charges

Charges +Q and +4Q are separated by distance d. Where is the zero-field point between them, measured from +Q?

Answer: d/3 from +Q
Formula used: kQ/x^2 = k(4Q)/(d - x)^2
  1. For cancellation between like charges, fields oppose between the charges.
  2. Set Q/x^2 = 4Q/(d - x)^2.
  3. d - x = 2x, so x = d/3.
JEE Main / IIT Main 8

JEE Main Electric Field Question 8: Null Point for Unequal Unlike Charges

Charges +Q at x = 0 and -4Q at x = d are fixed. Find the zero-field point on the x-axis.

Answer: At x = -d, outside on the side of +Q
Formula used: kQ/a^2 = k(4Q)/(d + a)^2
  1. For unlike unequal charges, the null point lies outside, near the smaller magnitude charge.
  2. Let the point be a distance a to the left of +Q.
  3. Then d + a = 2a, giving a = d and x = -d.
JEE Main / IIT Main 9

JEE Main Electric Field Question 9: Two Charges on Coordinate Axes

Equal positive charges q are placed at (a,0) and (0,a). Find the electric field at the origin in vector form.

Answer: -(kq/a^2)(i + j)
Formula used: Field due to a positive charge points away from that charge
  1. Charge at (a,0) gives field -(kq/a^2)i at the origin.
  2. Charge at (0,a) gives field -(kq/a^2)j.
  3. Net field is -(kq/a^2)(i + j).
JEE Main / IIT Main 10

JEE Main Electric Field Question 10: Regular Hexagon Symmetry

Six equal positive charges are placed at the vertices of a regular hexagon. What is the electric field at the centre?

Answer: Zero
Formula used: Opposite vertex fields cancel pairwise
  1. Each charge has an opposite charge at the same distance.
  2. The field vectors due to opposite vertices are equal and opposite.
  3. Three such pairs cancel, so net field is zero.
JEE Main / IIT Main 11

JEE Main Electric Field Question 11: Hexagon with One Charge Missing

In a regular hexagon of six equal charges, one charge is removed. Describe the field at the centre due to the remaining five charges.

Answer: It is equal and opposite to the field contribution of the removed charge
Formula used: E_six = E_removed + E_remaining = 0
  1. The full hexagon has zero field at the centre.
  2. Removing one charge removes its vector contribution.
  3. The remaining field must be the negative of the removed contribution.
JEE Main / IIT Main 12

JEE Main Electric Field Question 12: Logarithmic Slope for a Point Charge

For a point charge, what is the slope of a graph of log E versus log r?

Answer: -2
Formula used: E = kQ r^-2
  1. Taking logarithms gives log E = log(kQ) - 2 log r.
  2. The coefficient of log r is the slope.
  3. Therefore the slope is -2.
JEE Main / IIT Main 13

JEE Main Electric Field Question 13: Field Product for a Point Charge

For a fixed point charge, which product remains constant as r changes?

Answer: E r^2
Formula used: E = kQ/r^2
  1. Multiply both sides by r^2.
  2. E r^2 = kQ.
  3. Since Q is fixed, E r^2 is constant.
JEE Main / IIT Main 14

JEE Main Electric Field Question 14: Dipole Field Scaling

For a short dipole, if dipole moment doubles and distance from the centre doubles, what happens to the far axial field?

Answer: It becomes E/4
Formula used: E_axial = 2kp/r^3
  1. Field is proportional to p/r^3.
  2. New factor = 2/(2^3).
  3. New field is E/4.
JEE Main / IIT Main 15

JEE Main Electric Field Question 15: Short Dipole Numerical

A short dipole has p = 3 C m. Find its axial field at r = 3 m in terms of k.

Answer: 2k/9
Formula used: E_axial = 2kp/r^3
  1. Substitute p = 3 and r = 3.
  2. E = 2k(3)/(3^3).
  3. E = 6k/27 = 2k/9.
JEE Main / IIT Main 16

JEE Main Electric Field Question 16: Equatorial Direction of a Dipole

A dipole moment points along +x. What is the direction of electric field at a far equatorial point?

Answer: Along -x
Formula used: Equatorial dipole field is opposite to p
  1. At an equatorial point, transverse components cancel.
  2. Components along the dipole axis add opposite to p.
  3. Since p is +x, E is -x.
JEE Main / IIT Main 17

JEE Main Electric Field Question 17: Conductor Surface Condition

Why is the tangential component of electric field zero at the surface of a conductor in electrostatic equilibrium?

Answer: Because a tangential field would move free charges along the surface
Formula used: Electrostatic equilibrium requires no motion of free charges
  1. Conductors have mobile charges.
  2. A tangential electric field would exert tangential force.
  3. Charges would keep moving, contradicting equilibrium.
JEE Main / IIT Main 18

JEE Main Electric Field Question 18: Uniform Field Recognition

A field-line diagram shows straight, parallel, equally spaced lines. What kind of electric field is represented?

Answer: Uniform electric field
Formula used: Equal spacing means constant magnitude; parallel lines mean constant direction
  1. Parallel field lines indicate direction does not change.
  2. Equal spacing indicates strength does not change.
  3. Hence the field is uniform.
JEE Main / IIT Main 19

JEE Main Electric Field Question 19: Vector Force on Negative Charge

A charge q = -2 C is placed in electric field E = 3i - 4j N/C. Find the force vector.

Answer: -6i + 8j N
Formula used: F = qE
  1. Multiply each component of E by q = -2.
  2. F = -2(3i - 4j).
  3. F = -6i + 8j N.
JEE Main / IIT Main 20

JEE Main Electric Field Question 20: Charge from Force and Field

A particle experiences force 0.18 N in a field of magnitude 9.0 x 10^4 N/C. Find the magnitude of charge.

Answer: 2.0 microcoulomb
Formula used: |q| = F/E
  1. Use |q| = 0.18/(9.0 x 10^4).
  2. This equals 2.0 x 10^-6 C.
  3. Therefore the charge magnitude is 2.0 microcoulomb.
JEE Main / IIT Main 21

JEE Main Electric Field Question 21: Dipole Midpoint Field

Charges +q at x = -a and -q at x = +a form a dipole. Find the electric field direction at the origin.

Answer: Along +x
Formula used: Field due to +q is away from +q; field due to -q is toward -q
  1. At the origin, the +q contribution points right.
  2. The -q contribution also points right.
  3. Therefore net field is along +x.
JEE Main / IIT Main 22

JEE Main Electric Field Question 22: Negative Point Charge Direction

A charge -Q is placed at x = 0. At x = -a, what is the direction of electric field?

Answer: Along +x
Formula used: Field due to a negative source charge points toward the source
  1. The observation point is left of the negative charge.
  2. The field points toward the charge at the origin.
  3. Thus the direction is +x.
JEE Main / IIT Main 23

JEE Main Electric Field Question 23: External Field Cancellation

At a point P, a source charge produces 50 N/C east. A uniform external field of 50 N/C west is added. Find the net field at P.

Answer: Zero
Formula used: Collinear opposite vectors subtract
  1. Take east as positive.
  2. Net field = +50 - 50.
  3. So E_net = 0.
JEE Main / IIT Main 24

JEE Main Electric Field Question 24: Far Dipole Distance Change

For a short dipole, the far equatorial field at distance r is E. What is the field at distance 3r?

Answer: E/27
Formula used: E_equatorial = kp/r^3
  1. Dipole field varies inversely as r^3.
  2. Replacing r by 3r gives a factor 1/3^3.
  3. The field becomes E/27.
JEE Main / IIT Main 25

JEE Main Electric Field Question 25: Hollow Charged Conductor

A hollow conductor is in electrostatic equilibrium and has no charge inside its cavity. What is the field inside the conductor material?

Answer: Zero
Formula used: Electrostatic conductor condition: E = 0 inside conducting material
  1. Free charges in the conductor rearrange.
  2. The internal field inside the metal cancels.
  3. Hence E is zero inside the conducting material.

20 Original Difficult Questions

JEE Advanced / IIT Advanced Style Original Questions

Advanced-style original questions using multiple-correct, integer-type, assertion-reason, matching, paragraph-based and multi-step vector reasoning formats.

Multiple-correct | Difficult

JEE Advanced Original Question 1: Square Symmetry Multiple-Correct

Four identical positive charges are placed at the corners of a square. Which statements about the centre are correct?

  1. Net electric field is zero.
  2. Electric potential is positive.
  3. If one charge is removed, the remaining field is opposite to the removed charge contribution.
  4. Field lines intersect at the centre.
Complete solution:
  1. Opposite charges on diagonals give equal and opposite field vectors at the centre.
  2. Potential is scalar, so four positive contributions add.
  3. For the complete square E_total = 0, hence E_remaining = -E_removed if one charge is removed.
  4. Field lines never intersect, so option D is false.
Final answer: A, B and C
Key concept tested: Vector cancellation versus scalar addition
Integer-type | Moderate

JEE Advanced Original Question 2: Integer Null Point for Like Charges

Charges +Q and +9Q are separated by 40 cm. The zero-field point between them is x cm from +Q. Find x.

Complete solution:
  1. Let the point be x cm from +Q.
  2. Fields oppose between like charges, so Q/x^2 = 9Q/(40 - x)^2.
  3. 40 - x = 3x, giving x = 10 cm.
Final answer: 10
Key concept tested: Null point lies closer to the smaller like charge
Assertion-reason | Moderate

JEE Advanced Original Question 3: Assertion Reason on Field and Potential

Assertion: At the centre of a square with equal positive charges at all corners, electric field is zero. Reason: Electric potential at that centre is non-zero.

  1. Both true and the reason explains the assertion.
  2. Both true but the reason does not explain the assertion.
  3. Assertion true but reason false.
  4. Assertion false but reason true.
Complete solution:
  1. The assertion is true because field vectors cancel pairwise.
  2. The reason is also true because potential contributions add as scalars.
  3. However, non-zero potential is not the reason for zero field.
Final answer: B
Key concept tested: Do not mix electric field cancellation with potential addition
Matching-type | Moderate

JEE Advanced Original Question 4: Matching Field Dependence

Match: P point charge far field, Q short dipole axial far field, R short dipole equatorial direction, S electrostatic conductor interior.

  1. 1/r^2
  2. 1/r^3
  3. Opposite to p
  4. E = 0
Complete solution:
  1. A point charge follows inverse square law.
  2. A far short dipole follows inverse cube dependence.
  3. Equatorial field of a dipole is opposite to dipole moment.
  4. Inside a conductor in electrostatic equilibrium, the field is zero.
Final answer: P-1/r^2, Q-1/r^3, R-opposite to p, S-E = 0
Key concept tested: Rapid formula recognition across subtopics
Paragraph-based | Difficult

JEE Advanced Original Question 5: Hexagon with One Charge Removed

Six equal charges are placed on a regular hexagon. The field at the centre is zero. One charge at vertex A is removed. Determine the direction and magnitude form of the remaining field.

Complete solution:
  1. The full six-charge system gives E_total = 0 by symmetry.
  2. Let E_A be the field at the centre due to the charge at A.
  3. After removal, E_remaining = -E_A.
  4. The magnitude of E_A is kq/R^2, so the remaining magnitude is also kq/R^2.
Final answer: Opposite to the field due to the removed charge; magnitude kq/R^2 where R is the centre-to-vertex distance
Key concept tested: Use the zero sum of a symmetric full system
Multi-step numerical | Difficult

JEE Advanced Original Question 6: Dipole Equatorial Vector Derivation

Charges +q at (-a,0) and -q at (a,0) form a dipole. Find the electric field at P(0,b).

Complete solution:
  1. Field due to +q has direction from (-a,0) to (0,b).
  2. Field due to -q points from P toward (a,0).
  3. The y-components cancel and x-components add.
  4. The net field is 2kqa/(a^2 + b^2)^(3/2) along +x.
Final answer: E = [2kqa/(a^2 + b^2)^(3/2)] i
Key concept tested: Component cancellation on the perpendicular bisector
Multiple-correct | Difficult

JEE Advanced Original Question 7: Potential Zero but Field Non-Zero

At the midpoint of charges +Q and -Q separated by distance 2a, which statements are correct?

  1. Potential is zero.
  2. Electric field is zero.
  3. Electric field is directed from +Q to -Q.
  4. A positive charge placed there experiences force toward -Q.
Complete solution:
  1. Potential contributions +kQ/a and -kQ/a cancel.
  2. Field due to +Q and -Q point in the same direction at the midpoint.
  3. Therefore E is non-zero and points from +Q to -Q.
  4. A positive charge experiences force along E.
Final answer: A, C and D
Key concept tested: Scalar cancellation does not imply vector cancellation
Integer-type | Easy

JEE Advanced Original Question 8: Integer Ratio of Dipole Fields

For a short dipole, the axial field at distance r is n times the equatorial field at the same r. Find n.

Complete solution:
  1. E_axial = 2kp/r^3.
  2. E_equatorial = kp/r^3.
  3. The ratio E_axial/E_equatorial is 2.
Final answer: 2
Key concept tested: Axial field is twice equatorial field for a short dipole
Multiple-correct | Moderate

JEE Advanced Original Question 9: Conductor Equilibrium Multiple-Correct

For a conductor in electrostatic equilibrium, choose the correct statements.

  1. Electric field inside the conducting material is zero.
  2. Electric field at the surface has no tangential component.
  3. The conductor is an equipotential body.
  4. Excess charge is uniformly spread through the volume.
Complete solution:
  1. Free charges move until internal field is zero.
  2. A tangential surface field would cause charge motion.
  3. With no internal field, the conductor has constant potential.
  4. Excess charge resides on the surface, not uniformly through the volume.
Final answer: A, B and C
Key concept tested: Electrostatic equilibrium properties of conductors
Conceptual trap | Moderate

JEE Advanced Original Question 10: Field Line Logic Trap

A student draws two electrostatic field lines crossing near a high-field region and argues that the crossing only means the field is very strong. Is the drawing valid?

Complete solution:
  1. Strong field is represented by closely spaced field lines, not by intersecting lines.
  2. At an intersection, two tangents would exist at the same point.
  3. That would assign two different directions to electric field, which is impossible.
Final answer: No, the drawing is invalid
Key concept tested: Line density and line intersection have different meanings
Multi-step numerical | Difficult

JEE Advanced Original Question 11: Null Point Outside Unequal Unlike Charges

Charges +4Q at x = 0 and -Q at x = d are fixed. Find the null point on the x-axis.

Complete solution:
  1. For unequal unlike charges, the null point is outside on the side of the smaller magnitude charge.
  2. Let x > d. Distances from +4Q and -Q are x and x - d.
  3. Set 4/x^2 = 1/(x - d)^2.
  4. This gives x = 2d.
Final answer: x = 2d, on the side of -Q
Key concept tested: For unlike charges, the zero-field point is outside the segment
Vector/symmetry reasoning | Difficult

JEE Advanced Original Question 12: Triangle with One Negative Charge

At the vertices of an equilateral triangle, two charges are +q and one charge is -q. Find the field magnitude at the centre in terms of R, the centre-to-vertex distance.

Complete solution:
  1. If all three charges were +q, the centre field would be zero.
  2. Changing one +q to -q reverses that charge contribution, a net change of twice its original vector.
  3. Therefore the resultant has magnitude 2kq/R^2.
  4. The direction is toward the negative charge.
Final answer: 2kq/R^2, directed toward the negative charge
Key concept tested: Replace an asymmetric system by a symmetric reference system
Multi-step numerical | Difficult

JEE Advanced Original Question 13: Exact Versus Short Axial Formula

For a dipole with separation 2a, compare the exact axial field with the short-dipole axial field at r = 3a.

Complete solution:
  1. Exact axial field is E_exact = k(2pr)/(r^2 - a^2)^2.
  2. At r = 3a, E_exact = k(6pa)/(8a^2)^2 = 3kp/(32a^3).
  3. Short-dipole value is E_short = 2kp/(27a^3).
  4. The ratio is (3/32)/(2/27) = 81/64.
Final answer: E_exact/E_short = 81/64
Key concept tested: Short-dipole approximation improves only when r is much larger than a
Multiple-correct | Moderate

JEE Advanced Original Question 14: Force Direction Multiple-Correct

A uniform electric field points along +y. Which statements are correct?

  1. A proton experiences force along +y.
  2. An electron experiences force along -y.
  3. A negative ion experiences force opposite to the field.
  4. A neutral particle always has force qE with q = 1 C.
Complete solution:
  1. For positive charge, force is along electric field.
  2. For electron and negative ions, charge is negative, so force is opposite to E.
  3. A neutral particle has q = 0 for the simple electric force qE.
Final answer: A, B and C
Key concept tested: Sign of charge controls force direction
Paragraph-based | Difficult

JEE Advanced Original Question 15: Shielded Cavity Paragraph

A closed hollow conductor is placed in an external electrostatic field. No charge is kept inside the cavity. Discuss the field in the conductor material and inside the empty cavity.

Complete solution:
  1. Free charges rearrange on the conductor surfaces.
  2. The electric field inside conducting material becomes zero.
  3. With no charge inside a closed cavity, the electrostatic field in the cavity is also shielded in the ideal case.
  4. Surface charge redistribution supports these conditions.
Final answer: Field in the conductor material is zero; field in the empty closed cavity is also zero for electrostatic shielding
Key concept tested: Electrostatic shielding and conductor equilibrium
Matching-type | Difficult

JEE Advanced Original Question 16: Matching Zero Field and Zero Potential

Match the situation with the correct conclusion: P centre of equal positive charges on a square, Q midpoint of +Q and -Q, R point at infinity for a localized charge system.

  1. E = 0 but V not zero
  2. V = 0 but E not zero
  3. Both approach zero
Complete solution:
  1. At the square centre, field cancels by symmetry but positive potential adds.
  2. At the midpoint of equal unlike charges, potential cancels while fields add.
  3. Far away from a localized finite system, both field and potential tend to zero.
Final answer: P-E = 0 but V not zero; Q-V = 0 but E not zero; R-both approach zero
Key concept tested: Electric field and potential are related but not interchangeable
Multi-step numerical | Difficult

JEE Advanced Original Question 17: Three-Charge Vector Result

Charges +q at (a,0), +q at (-a,0), and -2q at (0,a) are fixed. Find electric field at the origin.

Complete solution:
  1. The fields at the origin due to charges at (a,0) and (-a,0) cancel.
  2. The -2q charge at (0,a) creates a field at the origin directed toward it, along +y.
  3. Its magnitude is k(2q)/a^2.
  4. Therefore E = 2kq/a^2 along +y.
Final answer: 2kq/a^2 along +y
Key concept tested: Cancel symmetric pair first, then evaluate the remaining vector
Assertion-reason | Moderate

JEE Advanced Original Question 18: Closed Loop Field Line Assertion

Assertion: Electrostatic field lines do not form closed loops. Reason: Electrostatic field is conservative.

  1. Both true and the reason explains the assertion.
  2. Both true but reason is unrelated.
  3. Assertion true but reason false.
  4. Assertion false but reason true.
Complete solution:
  1. Electrostatic field has zero curl in ordinary electrostatic situations.
  2. A closed field-line loop would imply circulation of electric field.
  3. That contradicts the conservative nature of electrostatic field.
Final answer: A
Key concept tested: Field-line geometry and conservative fields
Integer-type | Moderate

JEE Advanced Original Question 19: Integer Force Magnitude

A charge -2 microcoulomb is placed in E = (3i + 4j) x 10^5 N/C. Find the force magnitude in newton.

Complete solution:
  1. The field magnitude is 5 x 10^5 N/C.
  2. Force magnitude is |q|E = 2 x 10^-6 x 5 x 10^5.
  3. The result is 1 N.
Final answer: 1
Key concept tested: Magnitude uses |q|, direction uses sign of q
Conceptual trap | Difficult

JEE Advanced Original Question 20: Zero Field Conceptual Trap

At a point P, net electric field is zero due to several charges. Which conclusion is always valid?

  1. No charge is present anywhere nearby.
  2. Electric potential at P must be zero.
  3. Force on any charge placed at P is zero.
  4. Field lines must intersect at P.
Complete solution:
  1. E = 0 can occur by vector cancellation even when charges are nearby.
  2. Potential is scalar and may be non-zero.
  3. Since F = qE, the electric force on a charge placed at P is zero.
  4. Field lines do not intersect.
Final answer: C
Key concept tested: Meaning and limits of E = 0

10 Structured Questions

IB Physics Electric Field Questions

Structured explanation and data-based questions for electric field as force per unit charge, field-line interpretation, conductors and dipoles.

IB 1

IB Physics Question 1: Defining Electric Field

State what is meant by electric field strength at a point.

Answer: Electric field strength is the force per unit positive test charge placed at that point.
Explanation: The phrase positive test charge fixes the direction convention for the vector field.
IB 2

IB Physics Question 2: Unit Reasoning

Show why N C^-1 is a valid unit of electric field strength.

Answer: Since E = F/q, force is measured in newton and charge in coulomb, so the unit is N C^-1.
Explanation: The same physical quantity may also be written as V m^-1 in potential-gradient contexts.
IB 3

IB Physics Question 3: Field-Line Interpretation

A field-line diagram has closer lines near one electrode and wider spacing farther away. What does this show?

Answer: The electric field is stronger where the lines are closer together.
Explanation: IB questions often test interpretation of line density rather than only formula substitution.
IB 4

IB Physics Question 4: Data-Based Point Charge

A point charge produces 720 N/C at 0.50 m. Predict the field at 1.50 m.

Answer: 80 N/C
Explanation: Distance is tripled, and point-charge field varies as 1/r^2, so E becomes 720/9 = 80 N/C.
IB 5

IB Physics Question 5: Direction for Positive and Negative Charges

Compare the direction of electric field around isolated positive and negative charges.

Answer: It is outward from a positive charge and inward toward a negative charge.
Explanation: The direction is the direction of force on a small positive test charge.
IB 6

IB Physics Question 6: Conducting Shield

Explain why a closed conducting shell can shield its interior from an external electrostatic field.

Answer: Mobile charges redistribute on the conductor so that the field inside the conducting material, and ideally the empty cavity, becomes zero.
Explanation: This is an application of electrostatic equilibrium.
IB 7

IB Physics Question 7: Dipole Concept

What is an electric dipole and why is dipole moment a vector?

Answer: A dipole is two equal and opposite charges separated by distance; p = qd points from negative to positive charge.
Explanation: The direction matters because dipoles align and rotate in external electric fields.
IB 8

IB Physics Question 8: Superposition in Words

Describe how to find the net electric field at a point due to two charges.

Answer: Find each field separately at the point and add them as vectors.
Explanation: Adding magnitudes alone is correct only when the fields are already in the same direction.
IB 9

IB Physics Question 9: Force on Electron

An electron is placed in an electric field directed north. State the direction of electric force on it.

Answer: South.
Explanation: An electron has negative charge, so its force is opposite to the electric field direction.
IB 10

IB Physics Question 10: Moving Charge Context

Why is the electric field due to a moving charge not treated in the same simple way as a stationary point charge at introductory level?

Answer: Motion introduces magnetic effects and, at advanced level, relativistic considerations.
Explanation: School electrostatics first builds intuition using charges at rest, point charges, and dipoles.

20 Foundation Questions

ICSE / IGCSE Questions

Simple but conceptually strong short-answer, numerical and diagram-based questions on electric field direction, unit and field-line patterns.

ICSE-Level Electric Field Questions

ICSE 1

ICSE Electric Field Question 1: Simple Definition

Define electric field at a point.

Answer: It is the force experienced per unit positive test charge at that point.
Explanation: The word positive is important for deciding direction.
ICSE 2

ICSE Electric Field Question 2: Unit

Write one SI unit of electric field.

Answer: N/C.
Explanation: Electric field is force divided by charge.
ICSE 3

ICSE Electric Field Question 3: Positive Charge Pattern

How are field lines drawn around an isolated positive charge?

Answer: They radiate outward from the charge.
Explanation: A positive test charge is repelled by a positive source charge.
ICSE 4

ICSE Electric Field Question 4: Negative Charge Pattern

How are field lines drawn around an isolated negative charge?

Answer: They point inward toward the charge.
Explanation: A positive test charge would be attracted toward a negative source charge.
ICSE 5

ICSE Electric Field Question 5: Small Numerical

A charge of 2 C experiences a force of 10 N. Find the electric field.

Answer: 5 N/C.
Explanation: E = F/q = 10/2 = 5 N/C.
ICSE 6

ICSE Electric Field Question 6: Field Line Crossing

Why can two electric field lines not cross?

Answer: Crossing would show two directions of electric field at the same point.
Explanation: The electric field at one point has a unique direction.
ICSE 7

ICSE Electric Field Question 7: Stronger Field Region

What does closer spacing of electric field lines indicate?

Answer: A stronger electric field.
Explanation: Line density is a visual measure of relative field strength.
ICSE 8

ICSE Electric Field Question 8: Conductor Interior

What is the electric field inside a conductor in electrostatic equilibrium?

Answer: Zero.
Explanation: Free electrons move until the internal field is cancelled.
ICSE 9

ICSE Electric Field Question 9: Direction of Force

A positive charge is placed in an electric field directed east. What is the force direction?

Answer: East.
Explanation: A positive charge experiences force in the direction of electric field.
ICSE 10

ICSE Electric Field Question 10: Distance Effect

If the distance from a point charge becomes three times, how does field magnitude change?

Answer: It becomes one-ninth.
Explanation: Point charge field follows inverse square variation.

IGCSE-Level Electric Field Questions

IGCSE 1

IGCSE Electric Field Question 1: Meaning of Field

What is an electric field?

Answer: A region where an electric charge experiences a force.
Explanation: The field describes the effect of source charges on other charges.
IGCSE 2

IGCSE Electric Field Question 2: Positive Sphere Direction

A small positive test charge is near a positively charged sphere. Which way is it pushed?

Answer: Away from the sphere.
Explanation: Like charges repel, so the field direction is outward.
IGCSE 3

IGCSE Electric Field Question 3: Negative Sphere Direction

What is the direction of the electric field near a negatively charged sphere?

Answer: Toward the sphere.
Explanation: A positive test charge would be attracted to the negative sphere.
IGCSE 4

IGCSE Electric Field Question 4: Uniform Field Drawing

How are field lines shown in a uniform electric field?

Answer: Straight, parallel, and equally spaced.
Explanation: Equal spacing shows constant strength, and parallel lines show constant direction.
IGCSE 5

IGCSE Electric Field Question 5: Simple Force Calculation

A charge of 0.5 C is in an electric field of 20 N/C. Find the force.

Answer: 10 N.
Explanation: F = qE = 0.5 x 20 = 10 N.
IGCSE 6

IGCSE Electric Field Question 6: Charge Sign and Force

A negative charge is placed in a field pointing upward. What is the force direction?

Answer: Downward.
Explanation: Negative charges experience force opposite to the field.
IGCSE 7

IGCSE Electric Field Question 7: Strong and Weak Fields

On a field-line diagram, how can you identify the stronger field region?

Answer: Look for the region where field lines are closer together.
Explanation: Closer field lines represent larger field strength.
IGCSE 8

IGCSE Electric Field Question 8: Metal Shield

Why can a metal shell reduce electric field inside it?

Answer: Charges in the metal rearrange to oppose the external field.
Explanation: This is the basic idea of electrostatic shielding.
IGCSE 9

IGCSE Electric Field Question 9: Unit Recognition

Which unit is suitable for electric field strength: N, C, or N/C?

Answer: N/C.
Explanation: Electric field strength is force per unit charge.
IGCSE 10

IGCSE Electric Field Question 10: Comparing Points Near a Charge

Point A is closer to a charged sphere than point B. Where is the electric field usually stronger?

Answer: At point A.
Explanation: For a point-like charge distribution, field strength decreases with distance.

5 Case Studies

Case Study Questions

Case studies on point charge field, negative source charge, electric dipole, field lines and conductor in electrostatic equilibrium.

Case Study 1: Point Charge Field

A positive point charge +Q is fixed at the centre of a coordinate system. Students observe field directions and magnitudes at different distances.

Case 1.1

What is the direction of electric field at a point to the right of +Q?

Answer: To the right, away from +Q.
Explanation: A positive source charge produces radially outward electric field.
Case 1.2

If distance is doubled, what happens to field magnitude?

Answer: It becomes one-fourth.
Explanation: E = kQ/r^2 for a point charge.
Case 1.3

Does changing the test charge change the source field?

Answer: No, for an ideal small test charge.
Explanation: Electric field is determined by the source charge distribution.
Case 1.4

Which formula gives the field magnitude?

Answer: E = kQ/r^2.
Explanation: Here k = 1/(4 pi epsilon0).

Case Study 2: Negative Source Charge

A charge -Q is kept on an insulated stand. A small positive test charge is placed at nearby points to map the field.

Case 2.1

Do field lines start from -Q or end on -Q?

Answer: They end on -Q.
Explanation: Negative charges are sinks of electrostatic field lines.
Case 2.2

What force direction acts on a positive test charge?

Answer: Toward -Q.
Explanation: The electric field direction is defined using a positive test charge.
Case 2.3

What force direction acts on an electron at the same point?

Answer: Away from -Q.
Explanation: An electron is negative, so its force is opposite to electric field.
Case 2.4

Is the magnitude formula different from a positive charge of same |Q|?

Answer: No.
Explanation: The magnitude is k|Q|/r^2; the sign changes direction.

Case Study 3: Electric Dipole

Two charges -q and +q are separated by distance 2a. The dipole moment points from -q to +q.

Case 3.1

What is the dipole moment magnitude?

Answer: p = q x 2a.
Explanation: Dipole moment equals charge multiplied by separation.
Case 3.2

What is the axial short-dipole field magnitude?

Answer: E_axial = 2kp/r^3.
Explanation: This is valid when r is much greater than a.
Case 3.3

What is the equatorial short-dipole field magnitude?

Answer: E_equatorial = kp/r^3.
Explanation: It is half the axial magnitude at the same far distance.
Case 3.4

Which way is the equatorial field directed?

Answer: Opposite to p.
Explanation: Components perpendicular to the axis cancel.
Case 3.5

Why is dipole field faster-decaying than point charge field?

Answer: The total charge of a dipole is zero.
Explanation: The monopole contribution cancels, leaving inverse-cube behaviour far away.

Case Study 4: Field-Line Diagram

A student draws an electrostatic field-line diagram for two charges and uses arrows to show direction.

Case 4.1

What does the tangent to a field line represent?

Answer: Direction of electric field.
Explanation: The tangent gives the direction of force on a positive test charge.
Case 4.2

What does closer spacing of lines mean?

Answer: Stronger electric field.
Explanation: Line density represents relative field strength.
Case 4.3

Can two electrostatic field lines cross?

Answer: No.
Explanation: Crossing would assign two directions to one point.
Case 4.4

Do electrostatic field lines form closed loops?

Answer: No.
Explanation: They start on positive charges and end on negative charges or infinity.

Case Study 5: Conductor in Electrostatic Equilibrium

A hollow metal conductor is placed in an external electric field and allowed to reach electrostatic equilibrium.

Case 5.1

What is the field inside the conducting material?

Answer: Zero.
Explanation: Free charges rearrange until internal field cancels.
Case 5.2

How are field lines oriented at the surface?

Answer: Perpendicular to the surface.
Explanation: A tangential component would move charges.
Case 5.3

Where does excess charge reside?

Answer: On the surface.
Explanation: In electrostatic equilibrium, excess charge is not distributed through the bulk.
Case 5.4

What is the shielding idea?

Answer: The conductor reduces or cancels electrostatic field in protected regions.
Explanation: Redistributed surface charge opposes the applied field.
Case 5.5

Is the conductor an equipotential body?

Answer: Yes.
Explanation: With zero internal electric field, there is no potential difference within the conductor.

Last-Minute Revision

Final Revision Section

Use this section for quick NEET/JEE revision before solving mixed electric field questions.

Axial vs equatorial in one glance

  • Axial point lies on the dipole axis.
  • Equatorial point lies on the perpendicular bisector.
  • For a short dipole, E_axial = 2kp/r^3.
  • For a short dipole, E_equatorial = kp/r^3.
  • Equatorial direction is opposite to p.

Common mistakes checklist

  • Do not treat electric field as a scalar.
  • Do not reverse E when the test charge is negative; force reverses, not the source field.
  • Do not say field lines intersect at neutral points.
  • Do not use point-charge 1/r^2 variation for far dipole field.
  • Do not forget conductor interior field is zero only in electrostatic equilibrium.

Top 20 exam facts

  1. Electric field is force per unit positive test charge.
  2. Electric field is a vector.
  3. Positive charges produce outward field.
  4. Negative charges produce inward field.
  5. Point-charge field follows inverse square law.
  6. Force on positive charge is along E.
  7. Force on negative charge is opposite to E.
  8. Net field is vector sum of individual fields.
  9. Field lines never intersect.
  10. Field-line crowding means stronger field.
  11. Electrostatic field lines do not form closed loops.
  12. Conductor interior field is zero in electrostatic equilibrium.
  13. Field lines are normal to conductor surface.
  14. Dipole moment points from -q to +q.
  15. Dipole moment has unit C m.
  16. Far dipole field varies as 1/r^3.
  17. Axial short-dipole field is twice equatorial at same r.
  18. Equatorial dipole field is opposite to p.
  19. E can be zero while potential is non-zero.
  20. Potential can be zero while E is non-zero.

NEET/JEE quick revision points

  • For MCQs, draw arrows before substituting numbers.
  • For JEE vector questions, break fields into components.
  • For symmetry questions, first test if the complete system gives zero.
  • For null points, decide whether the charges are like or unlike before solving.
  • For conductor questions, look for the phrase electrostatic equilibrium.
  • For dipoles, check whether the point is axial or equatorial.
  • For short dipole formulas, confirm that r is much greater than a.

FAQ

Electric Field Concept FAQs

Short answers to common electric field concept questions asked by Class 12, NEET, JEE, IB, ICSE and IGCSE students.

What is electric field concept in Class 12 Physics?

Electric field concept explains force per unit positive test charge at a point and helps describe how charges influence the space around them.

What is the formula for electric field due to a point charge?

The magnitude is E = (1/4 pi epsilon0) Q/r^2, directed outward for positive Q and inward for negative Q.

What is the difference between axial and equatorial dipole field?

For a short dipole at the same distance, axial field has magnitude 2kp/r^3, while equatorial field has magnitude kp/r^3 and is opposite to dipole moment.

Why is electric field inside a conductor zero in electrostatic equilibrium?

Free charges in the conductor rearrange until the field produced by them cancels the applied field inside the conducting material.

Are these NEET and JEE electric field questions previous year questions?

No. They are original exam-style practice questions designed for NEET, JEE Main, and JEE Advanced preparation.

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