Electrostatic Potential and Capacitance

Take the positive charge as $+5\times10^{-8}\,\text{C}$ and the negative charge as $-3\times10^{-8}\,\text{C}$, separated by $16\,\text{cm}$. For potential to be zero,

$$\frac{5}{r_1}=\frac{3}{r_2}$$

because the potentials must cancel.

There are two possible points on the line:

1. Between the charges: if the point is $x$ cm from the positive charge, then it is $(16-x)$ cm from the negative charge.
$$\frac{5}{x}=\frac{3}{16-x}$$
$$5(16-x)=3x$$
$$80-5x=3x$$
$$8x=80 \Rightarrow x=10\,\text{cm}$$

2. On the extension beyond the negative charge: if the point is $x$ cm from the positive charge, then its distance from the negative charge is $(x-16)$.
$$\frac{5}{x}=\frac{3}{x-16}$$
$$5(x-16)=3x$$
$$5x-80=3x$$
$$2x=80 \Rightarrow x=40\,\text{cm}$$

So the points are 10 cm from the positive charge between the charges, and 40 cm from the positive charge on the side of the negative charge.

If the printed options do not include these exact values, the computed answer stands.

For a regular hexagon, the centre is at a distance equal to the side length from each vertex. So each charge is at distance

$$r=10\,\text{cm}=0.10\,\text{m}$$

from the centre.

Potential due to one charge:

$$V_1=\frac{1}{4\pi\varepsilon_0}\frac{q}{r} =9\times10^9\times\frac{5\times10^{-6}}{0.10}$$

$$V_1=9\times10^9\times5\times10^{-5}=4.5\times10^5\,\text{V}$$

There are 6 identical charges, so total potential is

$$V=6V_1=6\times4.5\times10^5=2.7\times10^6\,\text{V}$$

For two equal and opposite charges, every point on the perpendicular bisector is at equal distance from the two charges, so the potentials due to them cancel there. Hence it is an equipotential surface. In three dimensions, this is a plane perpendicular to the line joining the charges and passing through its midpoint.

The electric field is always normal to an equipotential surface. Therefore, at every point on this surface, the field is directed perpendicular to the surface.

Given a charged spherical conductor of radius $R=12\,\text{cm}=0.12\,\text{m}$ and charge $Q=1.6\times10^{-7}\,\text{C}$.

For a conductor:
- inside the conductor, $E=0$;
- just outside the surface,
$$E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{R^2}$$
- outside at distance $r$ from the centre,
$$E=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}$$

### (a) Inside the sphere
$$E=0$$

### (b) Just outside the sphere
$$E=9\times10^9\times\frac{1.6\times10^{-7}}{(0.12)^2}$$
$$(0.12)^2=0.0144$$
$$E=9\times10^9\times\frac{1.6\times10^{-7}}{0.0144} =9\times10^9\times1.111\times10^{-5} \approx 1.0\times10^5\,\text{N C}^{-1}$$

### (c) At 18 cm from the centre
Here $r=18\,\text{cm}=0.18\,\text{m}$.
$$E=9\times10^9\times\frac{1.6\times10^{-7}}{(0.18)^2}$$
$$(0.18)^2=0.0324$$
$$E=9\times10^9\times4.94\times10^{-6} \approx 4.4\times10^4\,\text{N C}^{-1}$$
The field is radially outward because the charge is positive.

If the book’s printed options are absent, these computed values are the correct ones.

For a parallel plate capacitor,

$$C=\varepsilon_0 K\frac{A}{d}$$

Initially the capacitance is $8\,\text{pF}$ with air, so the new capacitance changes by the factor

$$\text{factor} = K \times \frac{d}{d/2} = 6\times 2 = 12$$

Therefore,
$$C' = 12\times 8\,\text{pF} = 96\,\text{pF}$$

The question corresponds to Exercise 2.6(a): three capacitors of $9\,\text{pF}$ each are connected in series.

For series combination,
$$\frac{1}{C}=\frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$
With $C_1=C_2=C_3=9\,\text{pF}$,
$$\frac{1}{C}=\frac{1}{9}+\frac{1}{9}+\frac{1}{9}=\frac{3}{9}=\frac{1}{3}$$
Hence,
$$C=3\,\text{pF}$$