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: Understanding the Question:
We are given a conducting sphere of radius $r$ holding an isolated charge. The potential difference between its outer surface and an external space point located at a distance of $3r$ from the center is $V$. We need to find the absolute electric field strength ($E$) at that same external point.

Step 2: Key Formula or Approach:
For a charged conducting sphere, all internal charge resides on its outer surface, allowing it to be modeled as a point charge concentrated at the center for all external evaluations ($d \geq r$):
1. Electric Potential at a distance $d$ is: $V(d) = \frac{1}{4\pi\varepsilon_0}\frac{Q}{d}$ 2. Electric Field Intensity at a distance $d$ is: $E(d) = \frac{1}{4\pi\varepsilon_0}\frac{Q}{d^2}$ We use the potential difference to eliminate the unknown charge term $Q$.

Step 3: Detailed Explanation:
Let's define the electrostatic potentials at our two locations using $K = \frac{1}{4\pi\varepsilon_0}$:
At the surface ($d = r$):
$$V_{\text{surface}} = \frac{KQ}{r}$$ At the external point ($d = 3r$):
$$V_{\text{external}} = \frac{KQ}{3r}$$ The problem defines $V$ as the potential difference between these two points:
$$V = V_{\text{surface}} - V_{\text{external}} = \frac{KQ}{r} - \frac{KQ}{3r}$$ Factor out the common terms to simplify the fraction:
$$V = \frac{KQ}{r} \left( 1 - \frac{1}{3} \right) = \frac{KQ}{r} \left( \frac{2}{3} \right) = \frac{2KQ}{3r}$$ Let's rearrange this equation to express the point charge constant $KQ$ in terms of the potential difference $V$:
$$KQ = \frac{3rV}{2}$$ Now, write down the formula for the electric field intensity $E$ at that same external point ($d = 3r$):
$$E = \frac{KQ}{(3r)^2} = \frac{KQ}{9r^2}$$ Substitute our expression for $KQ$ into this electric field equation:
$$E = \frac{\left(\frac{3rV}{2}\right)}{9r^2} = \frac{3rV}{2 \times 9r^2} = \frac{3V}{18r} = \frac{V}{6r}$$ This simplifies cleanly to an intensity value of $\frac{V}{6r}$.

Step 4: Final Answer:
The electric field intensity at a distance of $3r$ from its center is $\frac{V}{6r}$, which matches option (A).