Question

A charged spherical conductor has radius $r$. The potential difference between its surface and a point at a distance of $3r$ from its centre is $V$. The electric field intensity at a distance of $3r$ from its centre is

• A. $\frac{V}{6r}$
• B. $\frac{V}{4r}$
• C. $\frac{V}{3r}$
• D. $\frac{V}{2r}$

Hint:

You can solve this quickly using a relative ratio approach: notice that the potential at $3r$ is exactly $\frac{1}{3}$ of the surface potential, making the difference $V = \frac{2}{3}V_{\text{surface}}$, which means $V_{\text{surface}} = 1.5V$. Since the electric field at any point can be written as $E = \frac{V_{\text{point}}}{d}$, the field at $3r$ is simply $\frac{V_{\text{external}}}{3r} = \frac{V_{\text{surface}}/3}{3r} = \frac{1.5V/3}{3r} = \frac{0.5V}{3r} = \frac{V}{6r}$.

Answer 1

The Correct Option is A

Step 1: Treat the sphere as a point charge.
Outside a charged conducting sphere of radius $r$, both potential and field behave as if all the charge $Q$ were at the centre. Write $K = \dfrac{1}{4\pi\varepsilon_0}.$
Step 2: Write the two potentials.
At the surface ($d = r$): $V_1 = \dfrac{KQ}{r}$. At the external point ($d = 3r$): $V_2 = \dfrac{KQ}{3r}.$
Step 3: Form the given potential difference.
$$V = V_1 - V_2 = \frac{KQ}{r} - \frac{KQ}{3r} = \frac{KQ}{r}\left(1 - \frac{1}{3}\right) = \frac{2KQ}{3r}.$$
Step 4: Solve for $KQ$.
Rearranging, $$KQ = \frac{3rV}{2}.$$
Step 5: Write the field at $3r$.
The field at distance $3r$ is $$E = \frac{KQ}{(3r)^2} = \frac{KQ}{9r^2}.$$
Step 6: Substitute $KQ$.
$$E = \frac{1}{9r^2}\times\frac{3rV}{2} = \frac{3V}{18r} = \frac{V}{6r}.$$
\[ \boxed{E = \frac{V}{6r}} \]