Question

A hollow charged metal sphere has radius 'R'. If the potential difference between its surface and a point at a distance '5R' from the centre is V, then magnitude of electric field Intensity at a distance '5R' from the centre of sphere is

• A. $\frac{V}{2R}$
• B. $\frac{V}{20R}$
• C. $10VR$
• D. $20VR$

Hint:

When relating $E$ and $V$ for point charges or spheres, finding their ratio is the most foolproof method. It quickly cancels out all the messy constants ($\frac{1}{4\pi\epsilon_0}$ and $q$) and leaves you with a simple algebraic relation in terms of $R$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given the potential difference ($V$) between the surface of a charged spherical shell and a point outside it at a distance of $5R$. We need to express the electric field ($E$) at that exact outside point in terms of $V$ and $R$.

Step 2: Key Formula or Approach:
The electric potential at the surface of a conducting sphere is $V_{\text{surface}} = \frac{1}{4\pi\epsilon_0}\frac{q}{R}$.
The electric potential at a point outside the sphere (distance $r = 5R$) is $V_{\text{outside}} = \frac{1}{4\pi\epsilon_0}\frac{q}{5R}$.
The electric field at a point outside the sphere is $E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$.

Step 3: Detailed Explanation:
First, write the expression for the potential difference $V$:
$$V = V_{\text{surface}} - V_{\text{outside}}$$ $$V = \frac{1}{4\pi\epsilon_0}\frac{q}{R} - \frac{1}{4\pi\epsilon_0}\frac{q}{5R}$$ Factor out the common terms:
$$V = \frac{1}{4\pi\epsilon_0}\frac{q}{R} \left(1 - \frac{1}{5}\right)$$ $$V = \frac{1}{4\pi\epsilon_0} \frac{4q}{5R}$$ Now, write the expression for the electric field $E$ at the distance $5R$:
$$E = \frac{1}{4\pi\epsilon_0}\frac{q}{(5R)^2}$$ $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{25R^2}$$ To find the relation between $E$ and $V$, take their ratio $\frac{E}{V}$:
$$\frac{E}{V} = \frac{\frac{1}{4\pi\epsilon_0}\frac{q}{25R^2}}{\frac{1}{4\pi\epsilon_0}\frac{4q}{5R}}$$ Cancel out $\frac{1}{4\pi\epsilon_0}$ and $q$:
$$\frac{E}{V} = \frac{\frac{1}{25R^2}}{\frac{4}{5R}} = \frac{1}{25R^2} \times \frac{5R}{4}$$ $$\frac{E}{V} = \frac{5R}{100R^2} = \frac{1}{20R}$$ Multiply both sides by $V$:
$$E = \frac{V}{20R}$$

Step 4: Final Answer:
The magnitude of electric field intensity is $\frac{V}{20R}$, matching option (B).