Question

A hollow metal sphere has a radius $r$. The potential difference between a point on its surface and at a point at a distance $3r$ from its centre is $V$. The electric intensity at the distance $3r$ from the centre of the sphere will be

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

Hint:

To remember this connection easily, look at the ratio between potential and field at any point outside a sphere: $E_x = \frac{V_x}{x}$. Here, the potential at $3r$ is $V_{3r} = \frac{kQ}{3r}$. Since $V = \frac{2kQ}{3r}$, we know $V_{3r} = \frac{V}{2}$. Plugging this into the ratio gives $E = \frac{V/2}{3r} = \frac{V}{6r}$ instantly!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem involves a hollow charged metallic sphere of radius $r$. For any closed conducting sphere, the electric field outside behaves as if the entire net charge $Q$ is concentrated at its center point.
We are given that the potential difference between the surface (at distance $r$) and an outer point (at distance $3r$) is $V$. We need to calculate the electric field intensity $E$ at that outer point $3r$.

Step 2: Key Formula or Approach:
1.

Electric Potential ($V_x$): The potential at any outer distance $x \geq r$ from the center of a charged sphere is: $$V_x = \frac{kQ}{x}$$ 2.

Electric Field Intensity ($E_x$): The electric field strength at an outer distance $x$ is: $$E_x = \frac{kQ}{x^2}$$ We will find the value of $kQ$ from the potential difference equation and substitute it into the field equation.

Step 3: Detailed Explanation:
Let's write down the absolute potential expressions at the two specified locations: Potential at the surface ($x = r$): $$V_{\text{surface}} = \frac{kQ}{r}$$ Potential at the outer point ($x = 3r$): $$V_{\text{outer}} = \frac{kQ}{3r}$$ The problem states that the magnitude of the potential difference between these two points is $V$: $$V = V_{\text{surface}} - V_{\text{outer}} = \frac{kQ}{r} - \frac{kQ}{3r}$$ Factor out $\frac{kQ}{r}$ to simplify the terms: $$V = \frac{kQ}{r} \left( 1 - \frac{1}{3} \right) = \frac{kQ}{r} \left( \frac{2}{3} \right)$$ $$V = \frac{2kQ}{3r}$$ Isolate the charge constant factor $kQ$ from this equation: $$kQ = \frac{3rV}{2}$$ Now, write the formula for the electric field intensity $E$ at the distance $x = 3r$: $$E = \frac{kQ}{(3r)^2} = \frac{kQ}{9r^2}$$ Substitute our derived expression for $kQ$ into this electric field equation: $$E = \frac{\left(\frac{3rV}{2}\right)}{9r^2}$$ Simplify the compound fraction terms carefully: $$E = \frac{3rV}{2 \cdot 9r^2} = \frac{3rV}{18r^2} = \frac{V}{6r}$$

Step 4: Final Answer:
The electric field intensity at the distance $3r$ is $\frac{V}{6r}$, which corresponds precisely to option (D).