Chapter 2: Electrostatic Potential and Capacitance (Electrostatics)

Put the positive charge at $x=0$ and the negative charge at $x=16$ cm. The potential is zero where $\frac{k(5\times 10^{-8})}{r_1}=\frac{k(3\times 10^{-8})}{r_2}$, that is $\frac{5}{r_1}=\frac{3}{r_2}$, where $r_1$, $r_2$ are the distances from the two charges.

Point between the charges (at $x$ cm): $r_1=x$, $r_2=16-x$.

$$\frac{5}{x}=\frac{3}{16-x}\Rightarrow 80-5x=3x\Rightarrow x=10\text{cm}$$

Point outside, beyond the negative charge: $r_1=x$, $r_2=x-16$.

$$\frac{5}{x}=\frac{3}{x-16}\Rightarrow 5x-80=3x\Rightarrow x=40\text{cm}$$

(No point exists beyond the positive charge, because there the larger positive charge is always nearer.)

At 10 cm and at 40 cm from the positive charge, on the side of the negative charge.

In a regular hexagon the distance from the centre to each vertex equals the side: $r=10$ cm $=0.1$ m. Potential is a scalar, so the six equal contributions add:

$$V=6\times \frac{kq}{r}=6\times \frac{9\times 10^9\times 5\times 10^{-6}}{0.1}=2.7\times 10^6\text{V}$$

$V=2.7\times 10^6$ V.

(a) Every point on the plane that is perpendicular to AB and passes through its midpoint is at equal distances from A and B. There $V=\frac{kq}{r}+\frac{k(-q)}{r}=0$. So this plane is an equipotential surface (at zero potential).

(b) The electric field is always normal to an equipotential surface. On this plane the field is perpendicular to the plane, parallel to AB, and points from A (positive) towards B (negative).

(a) The plane through the midpoint of AB, perpendicular to AB (V = 0). (b) Normal to the plane, in the direction from A to B.

(a) Inside a charged conductor the field is zero (all the charge is on the surface; a Gaussian surface inside encloses no charge).

(b) Just outside, the sphere acts like a point charge at its centre, with $r=0.12$ m:

$$E=\frac{kq}{r^2}=\frac{9\times 10^9\times 1.6\times 10^{-7}}{(0.12{)}^2}=\frac{1440}{0.0144}=1.0\times 10^5\text{N/C}$$

(c) At $r=0.18$ m:

$$E=\frac{1440}{(0.18{)}^2}=\frac{1440}{0.0324}\approx 4.4\times 10^4\text{N/C}$$

(a) 0. (b) $10^5$ N/C. (c) $4.4\times 10^4$ N/C (all directed radially outward outside).

$C_0=\frac{{\epsilon }_{0}A}{d}=8$ pF. New capacitance:

$$C=\frac{K{\epsilon }_{0}A}{d/2}=2K\cdot \frac{{\epsilon }_{0}A}{d}=2\times 6\times 8=96\text{pF}$$

96 pF.

(a) $\frac{1}{C}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}$, so $C=3$ pF.

(b) In series every capacitor carries the same charge, and here they are identical, so the 120 V is shared equally: $V=\frac{120}{3}=40$ V across each.

(Check: $Q=CV=3\text{pF}\times 120=360$ pC, and $360\text{pC}/9\text{pF}=40$ V.)

(a) 3 pF. (b) 40 V across each.

(a) $C=2+3+4=9$ pF.

(b) In parallel each capacitor has the full 100 V, and $Q=CV$:

• 2 pF: $Q=2\times 10^{-12}\times 100=2\times 10^{-10}$ C
• 3 pF: $Q=3\times 10^{-10}$ C
• 4 pF: $Q=4\times 10^{-10}$ C

(a) 9 pF. (b) $2\times 10^{-10}$ C, $3\times 10^{-10}$ C and $4\times 10^{-10}$ C.

$$C=\frac{{\epsilon }_{0}A}{d}=\frac{8.854\times 10^{-12}\times 6\times 10^{-3}}{3\times 10^{-3}}\approx 1.77\times 10^{-11}\text{F}=17.7\text{pF}$$

$$Q=CV=1.77\times 10^{-11}\times 100=1.77\times 10^{-9}\text{C}$$

One plate carries $+1.77\times 10^{-9}$ C and the other $-1.77\times 10^{-9}$ C.

$C\approx 17.7$ pF; $Q\approx 1.8\times 10^{-9}$ C on each plate.

The mica fills the whole gap, so the capacitance becomes $C^{\prime }=KC=6\times 17.7\approx 106$ pF.

(a) Supply connected: the battery keeps $V=100$ V. The charge rises to

$$Q^{\prime }=C^{\prime }V=1.06\times 10^{-10}\times 100\approx 1.06\times 10^{-8}\text{C}$$

six times the earlier charge (the battery pushes extra charge on).

(b) Supply disconnected first: the charge cannot change, $Q=1.77\times 10^{-9}$ C. The voltage falls:

$$V^{\prime }=\frac{Q}{C^{\prime }}=\frac{V}{K}=\frac{100}{6}\approx 16.7\text{V}$$

(a) V stays 100 V; charge becomes 6 times, about $1.06\times 10^{-8}$ C. (b) Charge stays $1.77\times 10^{-9}$ C; voltage drops to about 16.7 V.

$$U=\frac{1}{2}C{V}^2=\frac{1}{2}\times 12\times 10^{-12}\times (50{)}^2=1.5\times 10^{-8}\text{J}$$

$U=1.5\times 10^{-8}$ J.

Before: $Q=CV=600\times 10^{-12}\times 200=1.2\times 10^{-7}$ C and

$$U_1=\frac{1}{2}C{V}^2=\frac{1}{2}\times 600\times 10^{-12}\times (200{)}^2=1.2\times 10^{-5}\text{J}$$

After: the charge is shared by the two capacitors in parallel, $C^{\prime }=1200$ pF, and the charge is conserved:

$$V^{\prime }=\frac{Q}{C^{\prime }}=\frac{1.2\times 10^{-7}}{1200\times 10^{-12}}=100\text{V},U_2=\frac{1}{2}{C}^{\prime }{V}^{\prime 2}=\frac{1}{2}\times 1200\times 10^{-12}\times (100{)}^2=6\times 10^{-6}\text{J}$$

Energy lost $=U_1-U_2=1.2\times 10^{-5}-6\times 10^{-6}=6\times 10^{-6}$ J. It is lost as heat in the connecting wires and as electromagnetic radiation while the charge flows.

$6\times 10^{-6}$ J is lost (half the original energy).