Electrostatics — Practice

Question 1

Compare the axial and equatorial fields of a dipole at the same distance, and state the potential at each point.

Accepted Answer

Axial $E = 2kp/r^{3}$, equatorial $E = kp/r^{3}$, so the axial field is exactly twice as large. Axial $V = kp/r^{2}$; equatorial $V = 0$, since $\cos	heta = 0$ there. The equatorial point therefore has zero potential with a non-zero field.

Question 2

A charged conducting shell of radius $R$ carries charge $Q$. Give the field and potential at $r = R/2$.

Accepted Answer

$E = 0$, since a Gaussian sphere inside encloses no charge. But $V 
eq 0$: zero field means constant potential, so the interior holds the surface value $V = kQ/R$ throughout.

Question 3

Three capacitors of 2, 3 and 6 $\mu$F are connected in series, and then in parallel, across 12 V. Find each equivalent capacitance and the energy stored.

Accepted Answer

Series: $1/C = 1/2 + 1/3 + 1/6 = 1$, so $C_{eq} = 1\ \mu$F and $U = 	frac{1}{2}(10^{-6})(144) = 7.2	imes10^{-5}$ J. Parallel: $C_{eq} = 11\ \mu$F and $U = 7.92	imes10^{-4}$ J. The series equivalent is smaller than the smallest member — the reverse of resistors.

Question 4

A point charge $q$ sits at the centre of a cube. Find the flux through one face. Then find it if the charge is moved to a corner.

Accepted Answer

At the centre the total flux $q/\varepsilon_0$ is shared equally by six faces, giving $q/6\varepsilon_0$ each. At a corner the charge is shared between eight cubes, so this cube encloses $q/8$. The three faces touching the corner have the field running along them and carry zero flux, so the other three share $q/8\varepsilon_0$, giving $q/24\varepsilon_0$ each.

Question 5

Charges $+q$, $+q$ and $-q$ sit at $x = 0$, $a$ and $2a$. Find the net force on the middle charge.

Accepted Answer

The charge at the origin repels it along $+x$ with magnitude $kq^{2}/a^{2}$; the negative charge at $2a$ attracts it, also along $+x$, with the same magnitude. The net force is $2kq^{2}/a^{2}$ along $+x$ — showing that a repulsion and an attraction can push the same way, so signs must be tracked geometrically rather than by formula.

Question 6

Find the point on the axis of a uniformly charged ring of radius $R$ where the electric field is maximum.

Accepted Answer

$E = kQx/(x^{2}+R^{2})^{3/2}$ is zero at the centre by symmetry and tends to zero far away, so it must peak in between. Differentiating and setting $dE/dx = 0$ gives $R^{2} = 2x^{2}$, so $x = R/\sqrt{2}$.

Question 7

A capacitor is charged to 100 V and the battery is then disconnected. A dielectric of constant $K = 4$ is inserted. Find the new voltage and the ratio of final to initial energy.

Accepted Answer

With the battery disconnected, $Q$ is fixed. Capacitance quadruples, so $V = Q/4C$ falls to 25 V. Since $U = Q^{2}/2C$, the energy falls to one quarter. The missing energy is the work the capacitor did pulling the dielectric into the gap.

Question 8

Explain why the electric field of an infinite charged sheet does not depend on the distance from the sheet.

Accepted Answer

Each element's contribution does weaken as $1/r^{2}$, but moving further away brings a proportionally larger area of sheet into view. For a genuinely infinite sheet the two effects cancel exactly, leaving $\sigma/2\varepsilon_0$ at every distance. Real sheets show this only while the distance stays small compared with the sheet's own dimensions.

Question 9

A capacitor $C_1$ charged to $V_1$ is connected to an uncharged $C_2$. Find the common potential and the energy lost, and explain where it goes.

Accepted Answer

Charge conservation gives $V = C_1V_1/(C_1+C_2)$. The energy lost is $C_1C_2V_1^{2}/2(C_1+C_2)$, appearing as heat in the wires and as radiation. Remarkably the loss does not depend on the wire's resistance: halving $R$ halves the time but doubles the current, dissipating the same total.

Question 10

Explain why the field inside a conductor in electrostatic equilibrium must be exactly zero, and list three consequences.

Accepted Answer

If any field existed, the free charges would feel a force and move, so the situation would not be static. Consequences: all excess charge sits on the surface, since interior charge would produce an interior field by Gauss's law; the whole conductor is one equipotential, since moving charge within it needs no work; and any cavity is shielded from external fields, which is electrostatic shielding.

Question 11

A solid conducting sphere of radius $R$ carries total charge $Q$. Find $E$ and $V$ both inside and outside, and explain why the potential is not zero inside even though the field is.

Accepted Answer

Outside, the sphere behaves as a point charge at its centre: $E = kQ/r^{2}$ and $V = kQ/r$. Inside, all the charge sits on the surface, so a Gaussian sphere encloses nothing and $E = 0$; since $E = -dV/dr$, a zero field means $V$ is constant, and continuity at the surface fixes that constant at $kQ/R$. Potential measures the work needed to bring a test charge in from infinity, and that journey does work on the way to the surface even though the final leg inside does none.

Question 12

A parallel plate capacitor of capacitance $C_0$ is charged to $V_0$ and then disconnected from the battery, after which a dielectric slab of constant $K$ is inserted to fill the gap. Find the new potential difference and stored energy. Repeat with the battery left connected, and say in which case the slab is drawn in.

Accepted Answer

Disconnected, the charge $Q = C_0V_0$ is fixed. Capacitance becomes $KC_0$, so $V = V_0/K$ and $U = Q^{2}/2C = U_0/K$: the stored energy falls. Connected, the voltage is fixed instead. Capacitance becomes $KC_0$, so $Q$ rises to $KC_0V_0$ and $U = 	frac12 KC_0V_0^{2} = KU_0$: the stored energy rises, but the battery supplies $(K-1)C_0V_0^{2} = 2(K-1)U_0$, which is twice the increase. The slab is pulled in either way — at constant charge because the system's energy falls, and at constant voltage because the battery delivers twice what the field energy gains, the surplus doing the pulling.

Practice — Electrostatics

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