Electrostatic Potential and Capacitance — NCERT Solutions

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 equal and opposite charges, the potential at a point is the algebraic sum of the potentials due to each charge. On the perpendicular bisector of the line joining the two charges, every point is at the same distance from $+2\,\mu\text{C}$ and $-2\,\mu\text{C}$. Since the charges are equal in magnitude and opposite in sign, their potentials cancel, so the net potential is zero. Hence this is an equipotential surface.

On an equipotential surface, the potential is the same at every point, so there is no potential difference along the surface. Therefore the electric field cannot have any component along the surface. Hence the electric field at every point on this surface must be normal to the equipotential 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,
$$C' = 12\times 8\,\text{pF} = 96\,\text{pF}$$

For capacitors in series,\n$$\frac{1}{C} = \frac{1}{C_1}+\frac{1}{C_2}+\frac{1}{C_3}$$\nHere each capacitor is $9\,\text{pF}$, so\n$$\frac{1}{C} = \frac{1}{9}+\frac{1}{9}+\frac{1}{9} = \frac{3}{9} = \frac{1}{3}$$\nThus,\n$$C = 3\,\text{pF}$$