coulombs law intuitive Electric Charge and Electric Dipole

Solution: Use F = kq₁q₂/r². With r = 0.30 m, F = 9 × 10⁹ × 2 × 10⁻⁶ × 3 × 10⁻⁶ / 0.09 = 0.60 N, repulsive.

Solution: Electrons moved from the glass to the silk. Protons stayed bound in nuclei, so the glass has electron deficiency and becomes positive.

Solution: n = Q/e = 1.6 × 10⁻⁶ / 1.6 × 10⁻¹⁹ = 10¹³ electrons. Positive charge means these electrons are missing.

Solution: Since F ∝ 1/r², doubling distance makes force F/4. This conclusion is independent of the signs of the charges.

Solution: Here 2a = 0.04 m. p = q(2a) = 5 × 10⁻⁶ × 0.04 = 2.0 × 10⁻⁷ C m, directed from −q to +q.

Solution: The rod is not a point charge. It must be divided into small elements dq and the contributions must be integrated.

Solution: Fmedium = Fvacuum/K = Fvacuum/80. The dielectric reduces the electric interaction by polarising between the charges.

Solution: Force is proportional to q₁q₂. Doubling one charge doubles the product, so the new force is 1.8 N.

Solution: SI unit is C m. Since p = q × length, dimensions are [A T] [L] = [A L T].

Solution: The total positive charge of protons equals the total negative charge of electrons, so algebraic charge sum is zero.

Solution: No. Isolated free charge is quantised as Q = ne, where n is an integer. 2.5e is not allowed for a body.

Solution: It settles on the outer surface. Inside a conductor at electrostatic equilibrium, the electric field is zero and excess charge cannot remain in the bulk.

Solution: Total charge is +4 μC and identical spheres share equally, so each gets +2 μC.

Solution: The wall molecules polarise. Opposite bound charge appears slightly closer to the balloon than like bound charge, creating net attraction.

Solution: τ = pE sinθ, so torque is zero at θ = 0° and 180°. The stable orientation is p→ parallel to E→.

Solution: For like charges, zero field lies between them. Let x be distance from 4 μC: 4/x² = 9/(1 − x)², so 2/x = 3/(1 − x), giving x = 0.40 m.

Solution: Charge product becomes 4 times and r² becomes 9 times. New force = 4F/9.

Solution: Electrostatic force dominates because Fe/Fg = ke²/(Gmₑ²) is about 4 × 10⁴² for two electrons.

Solution: For r much larger than separation, Eaxial = 2kp/r³. Direction is along p→ on the positive axial side.

Solution: Eequatorial = −kp/r³. The minus sign says the field is opposite to p→.

Solution: The perpendicular components cancel by symmetry. The components along the dipole axis add towards −Q.

Solution: Magnitude F = 9 × 10⁹ × 10⁻¹² / 10⁻² = 0.9 N. Since signs are opposite, the force is attractive.

Solution: Induction. A nearby charged body redistributes charge; earthing lets one sign escape or enter; removing earth leaves net charge.

Solution: The force becomes one-sixth of the original value: F = F₀/K.

Solution: Charge −8e means eight extra electrons compared with neutrality.

Solution: p = q(2a). Doubling q and halving separation leaves the product unchanged, so p remains 3 C m.

Solution: Electric charge has two signs; mass has only one sign in classical mechanics. Like charges repel, unlike charges attract.

Solution: Perpendicular vector sum gives F = √(3² + 4²) = 5 N at tan⁻¹(4/3) north of east.

Solution: F ∝ q/r² for one charge changed. New factor = 5/(√5)² = 1, so force is unchanged.

Solution: Axial magnitude is 2kp/r³ and equatorial magnitude is kp/r³, so axial is twice as strong.

Solution: The two attractive forces each have magnitude kq²/a² and angle 60° between them. Resultant = 2F cos30° = √3 kq²/a² along the angle bisector.

Solution: Let x be distance from +Q. kQq/x² = k4Qq/(L − x)². Thus (L − x)/x = 2, so x = L/3.

Solution: The two adjacent charges contribute kq/a² along perpendicular sides; diagonal charge contributes kq/(2a²) along the diagonal. Resolve along diagonal: E = √2 kq/a² + kq/(2a²).

Solution: a = F/m = kQq/(mr²), away from Q. This is instantaneous because r changes after release.

Solution: Initial product magnitude is 16. Total charge is 6 μC, so final charge on each is 3 μC and product is 9. Ratio final/initial = 9/16.

Solution: E = kq[(r − a)⁻² − (r + a)⁻²] = kq(4ar)/(r² − a²)² = k(2pr)/(r² − a²)².

Solution: For unlike charges, zero field is outside, on the side of the smaller magnitude. Let point be x = −d. Q/d² = 4Q/(L + d)², so L + d = 2d and d = L.

Solution: Equal vectors symmetrically separated by 120° add to zero, so net force is zero.

Solution: By symmetry, forces from opposite corners cancel pairwise. Net force at the centre is zero.

Solution: F ∝ 1/r². If F' = F/9, then r'² = 9d² and r' = 3d.

Solution: No. Force depends on electric field, not potential alone. Potential can be zero while field is non-zero, as with equal unlike charges at midpoint.

Solution: r = √(kq₁q₂/F) = √(9 × 10⁹ × 100 × 10⁻¹² / 18) = √0.05 ≈ 0.224 m.

Solution: At midpoint, fields due to +q and −q of the dipole point from + to −, so they add. Force is qtest times that net field towards − charge.

Solution: Since a²/r² = 1/100 is small, terms like (1 − a²/r²)⁻² can be approximated as 1. Use short dipole results.

Solution: Horizontal components cancel and vertical components add. For positive source charges, force is away from the x-axis along +y if the test charge is above.

Solution: F = kq(Q − q)/r². Maximise f(q) = qQ − q². df/dq = Q − 2q = 0, so q = Q/2 and the split is equal.

Solution: Two side forces are F = kq²/a² perpendicular; their resultant is √2F along diagonal. Diagonal force is kq²/(2a²) along same diagonal. Net = (√2 + 1/2)kq²/a².

Solution: Exact field is −kp/(r² + a²)³ᐟ². For r ≫ a, write (r² + a²)³ᐟ² = r³(1 + a²/r²)³ᐟ². Leading term is −kp/r³.

Solution: Along x, attraction towards nearer charge grows and equilibrium is unstable. Along y, horizontal components cancel and vertical components pull back toward origin, giving stable small displacement.

Solution: Forces have magnitudes 2kq²/a² and 3kq²/a² with angle 60° between them. Resultant magnitude = (kq²/a²)√(4 + 9 + 12 cos60°) = √19 kq²/a².

Solution: Resolve Coulomb force components and set net force zero: kQq r→/r³ + m g→ + constraint forces = 0. For a free bead, vertical component must equal mg and horizontal component must vanish.

Solution: Forces on +q and −q are equal and opposite only in a uniform field. If E varies over the separation, magnitudes differ, producing net force.

Solution: Between unlike charges, field due to +q is away from +q and field due to −q is toward −q. Both point from +q to −q, so they add.

Solution: No. The monopole term vanishes, but a dipole term may remain. A neutral dipole has far field proportional to 1/r³.

Solution: Only if their signs are opposite so the forces on q₀ oppose. If both signs relative to q₀ are same and same side, forces are in same direction.

Solution: Replace ε₀ by ε = Kε₀. The interaction coefficient becomes k/K and force reduces by factor K.

Solution: R² = F² + 4F² + 2F·2F cos120° = 5F² − 2F² = 3F². Hence R = √3 F.

Solution: Exact/short = r⁴/(r² − a²)² = 625/576 ≈ 1.085. Short formula is about 8.5% low.

Solution: Use vector components with signs before rounding. Subtracting nearly equal large numbers can amplify rounding error.

Solution: Coulomb force from each source is proportional to that source charge, and superposition adds forces without source-source modification in electrostatics.

Solution: F = 8.99 × 10⁹ × 18 × 10⁻¹⁸ / 0.040 = 4.05 × 10⁻⁶ N. The force is attractive.

Solution: The force on a positive test charge has both magnitude and direction. Since E = F/qtest, field inherits the vector direction of force.

Solution: Q/e = −3. The body has three excess electrons.

Solution: Both are inverse-square central-force laws acting along the line joining the two interacting bodies.

Solution: Electric force may attract or repel because charge has two signs; gravitational force between ordinary masses is attractive.

Solution: p = qd = 2.0 × 10⁻⁹ × 5.0 × 10⁻³ = 1.0 × 10⁻¹¹ C m.

Solution: Dipole field varies as 1/r³. Doubling r reduces the field to 1/8.

Solution: Fields are vectors. Components must be resolved and added; magnitudes add only when the field vectors are collinear and same direction.

Solution: It is K = ε/ε₀, the ratio of permittivity of a medium to vacuum permittivity, and it measures how strongly the medium reduces electric interaction.

Solution: Friction transfers electrons between hair and comb; the paper bits are then attracted mainly by induction/polarisation.

Solution: It consists of equal and opposite charges. Net charge is +q − q = 0, although charge separation gives non-zero dipole moment.

Solution: At the equatorial point, vertical components cancel. Horizontal components from both charges point toward the negative charge, opposite to p→.

Solution: F = qE = 2 × 10⁻⁶ × 300 = 6.0 × 10⁻⁴ N in the direction of the field.

Solution: Molecules polarise, creating bound charge whose field partly opposes the original applied field.

Solution: The separation tends to zero and charge tends to infinity while p stays finite. Real dipoles have finite separation and finite charges.