"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.

Practice — "Electrostatics"

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