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
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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 $R$.
Step 1: Understand the question.
A hollow charged metal sphere has radius $R$. The potential difference between its surface and a point at distance $5R$ from the centre is $V$. We must find the electric field at that distance $5R$.
Step 2: Write the two potentials.
The potential at the surface ($r = R$) and at the outside point ($r = 5R$) are:
\[ V_{\text{surface}} = \frac{1}{4\pi\epsilon_0}\frac{q}{R}, \qquad V_{5R} = \frac{1}{4\pi\epsilon_0}\frac{q}{5R} \]
Step 3: Write the potential difference.
\[ V = V_{\text{surface}} - V_{5R} = \frac{1}{4\pi\epsilon_0}\frac{q}{R}\left(1 - \frac{1}{5}\right) = \frac{1}{4\pi\epsilon_0}\frac{4q}{5R} \]
Step 4: Write the field at $5R$.
Outside the sphere the field is like that of a point charge:
\[ E = \frac{1}{4\pi\epsilon_0}\frac{q}{(5R)^2} = \frac{1}{4\pi\epsilon_0}\frac{q}{25R^2} \]
Step 5: Divide field by potential difference.
The $\tfrac{1}{4\pi\epsilon_0}$ and $q$ cancel:
\[ \frac{E}{V} = \frac{\dfrac{q}{25R^2}}{\dfrac{4q}{5R}} = \frac{1}{25R^2}\times \frac{5R}{4} = \frac{5R}{100R^2} = \frac{1}{20R} \]
Step 6: Solve for $E$.
Multiply both sides by $V$:
\[ E = \frac{V}{20R} \]
This matches option (2).
\[ \boxed{E = \frac{V}{20R}} \]