Electrostatics — Practice Questions with Answers

Coulomb's law states that the electrostatic force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them, acting along the line joining them.

$\vec{F} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1q_2}{r^2}\hat{r}$, where $\epsilon_0$ is the permittivity of free space, and $\hat{r}$ is the unit vector along the line joining the charges.

Electric field intensity at a point is defined as the force experienced per unit positive test charge placed at that point: $\vec{E} = \vec{F}/q_0$.

For a point charge $Q$, the force on a small test charge $q_0$ at distance $r$ is $F = \dfrac{kQq_0}{r^2}$. Dividing by $q_0$:

$E = \dfrac{kQ}{r^2} = \dfrac{1}{4\pi\epsilon_0}\dfrac{Q}{r^2}$

Electric potential at a point is the work done per unit positive test charge in bringing it from infinity to that point, without acceleration: $V = W/q_0$.

For a uniform field $E$, the potential difference between two points separated by distance $d$ (along the field direction) is $V = Ed$, or more generally, $E = -\dfrac{dV}{dr}$ — the field equals the negative gradient of potential.

Gauss's law states that the total electric flux through any closed surface equals the net charge enclosed divided by $\epsilon_0$: $\oint\vec{E}\cdot d\vec{A} = \dfrac{q_{enc}}{\epsilon_0}$.

Its significance lies in dramatically simplifying electric field calculations for charge distributions with high symmetry (spherical, cylindrical, or planar) — by choosing an appropriate "Gaussian surface" matching the symmetry, the field can often be found algebraically, without the need for direct, often complex integration used in Coulomb's law-based calculations.

An electric dipole is a pair of equal and opposite point charges separated by a small distance. Dipole moment $p = q\times2a$ (where $2a$ is the separation), a vector pointing from the negative to positive charge.

Electric field on the axial line, at distance $r$ from the dipole's centre (for $r \gg a$): $E_{axial} = \dfrac{2kp}{r^3}$

A conductor contains free (mobile) charge carriers. If any net electric field existed within the conductor, it would exert a force on these free charges, causing them to move (a current).

In electrostatic equilibrium, by definition, there is no ongoing charge movement — the free charges have already redistributed themselves (moving to the conductor's surface) in such a way that their own electric field exactly cancels any external field within the conductor's interior, resulting in zero net internal field. This redistribution happens essentially instantaneously for good conductors.

A capacitor is a device that stores electric charge (and energy) using two conductors separated by an insulator, developing a potential difference proportional to the stored charge: $Q = CV$.

For a parallel plate capacitor with plate area $A$ and separation $d$: the field between plates is $E = \sigma/\epsilon_0 = Q/(A\epsilon_0)$, and $V=Ed = Qd/(A\epsilon_0)$.

$C = \dfrac{Q}{V} = \dfrac{\epsilon_0A}{d}$

In series, the same charge $Q$ appears on each capacitor, with total voltage split across them: $V = V_1 + V_2 + V_3 + \cdots$

Since $V_i = Q/C_i$ for each capacitor: $\dfrac{Q}{C_{eq}} = \dfrac{Q}{C_1} + \dfrac{Q}{C_2} + \dfrac{Q}{C_3} + \cdots$

Dividing by $Q$: $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} + \cdots$

Electrostatic shielding refers to the phenomenon where a hollow conductor shields its interior from any external electric field — since the field inside a conductor (including any hollow cavity within it) in electrostatic equilibrium is always zero, external fields cannot penetrate into the enclosed cavity.

Application: Sensitive electronic equipment is often enclosed within a conducting metal case (a "Faraday cage") to shield it from external electromagnetic interference, and it's also why occupants inside a car (with a largely metallic body) are relatively safe from external electric effects, such as during a lightning strike.

As charge is built up on a capacitor, small work is done moving each additional charge $dq$ against the existing potential $V=q/C$: $dW = V\,dq = \dfrac{q}{C}dq$

Total work (= stored energy) to charge from 0 to $Q$: $U = \displaystyle\int_0^Q \dfrac{q}{C}dq = \dfrac{Q^2}{2C}$

This can also be written as $U = \frac{1}{2}CV^2 = \frac{1}{2}QV$ (using $Q=CV$).

$F = \dfrac{kq_1q_2}{r^2} = \dfrac{9\times10^9\times4\times10^{-6}\times9\times10^{-6}}{(0.3)^2}$

Numerator: $9\times4\times9 = 324$, so $324\times10^{-3} = 0.324$

$F = \dfrac{0.324}{0.09} = \mathbf{3.6\text{ N}}$ (repulsive, since both charges are positive)

$E = \dfrac{kQ}{r^2} = \dfrac{9\times10^9\times5\times10^{-6}}{(0.2)^2}$

$E = \dfrac{45\times10^3}{0.04} = \mathbf{1.125\times10^6\text{ N/C}}$

$C = \dfrac{\epsilon_0A}{d} = \dfrac{8.85\times10^{-12}\times0.02}{2\times10^{-3}}$

$C = \dfrac{1.77\times10^{-13}}{2\times10^{-3}} = \mathbf{8.85\times10^{-11}\text{ F} = 88.5\text{ pF}}$

Parallel: $C_{eq} = C_1+C_2+C_3 = 2+3+6 = \mathbf{11\,\mu\text{F}}$

Series: $\dfrac{1}{C_{eq}} = \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6} = \dfrac{3}{6}+\dfrac{2}{6}+\dfrac{1}{6} = \dfrac{6}{6} = 1$

$C_{eq} = \mathbf{1\,\mu\text{F}}$

(a) Charge: $Q = CV = 10\times10^{-6}\times100 = 1\times10^{-3}\text{ C} = 1\text{ mC}$

(b) Energy stored: $U = \frac{1}{2}CV^2 = \frac{1}{2}(10\times10^{-6})(100)^2 = \frac{1}{2}(10\times10^{-6})(10000)$

$U = \frac{1}{2}\times0.1 = \mathbf{0.05\text{ J}}$

$p = q\times d = 2\times10^{-6}\times4\times10^{-2}$

$p = \mathbf{8\times10^{-8}\text{ C·m}}$, directed from the negative to the positive charge.

Work done = charge × potential difference (in the direction of decreasing potential, work done by the field): $W = q(V_1-V_2)$

$W = 2\times10^{-6}\times(100-40) = 2\times10^{-6}\times60$

$W = \mathbf{1.2\times10^{-4}\text{ J} = 120\,\mu\text{J}}$

(a) New capacitance: $C' = KC = 4\times5 = 20\,\mu\text{F}$

(b) New charge (voltage stays at 12 V, since still connected to battery): $Q' = C'V = 20\times10^{-6}\times12$

$Q' = \mathbf{2.4\times10^{-4}\text{ C} = 240\,\mu\text{C}}$

Force between any two charges: $F = \dfrac{kq^2}{r^2} = \dfrac{9\times10^9\times(1\times10^{-6})^2}{(0.1)^2} = \dfrac{9\times10^9\times10^{-12}}{0.01} = 0.9\text{ N}$

Each charge experiences two equal forces of 0.9 N from the other two, at 60° to each other (angle of equilateral triangle). Resultant of two equal forces $F$ at angle $\theta$: $F_{net} = 2F\cos(\theta/2)$

$F_{net} = 2(0.9)\cos(30°) = 1.8\times0.866 \approx \mathbf{1.56\text{ N}}$, directed away from the triangle's centre.

By Gauss's law, total flux through the entire (closed) cube surface: $\Phi_{total} = \dfrac{q}{\epsilon_0} = \dfrac{8\times10^{-6}}{8.85\times10^{-12}} \approx 9.04\times10^5\text{ N·m}^2/\text{C}$

By symmetry, this total flux is distributed equally among all 6 faces of the cube (since the charge is at the exact centre):

$\Phi_{one\,face} = \dfrac{9.04\times10^5}{6} \approx \mathbf{1.51\times10^5\text{ N·m}^2/\text{C}}$