Question:
A spherical shell of radius \(R\) has charge \(Q\) uniformly distributed over its surface. The work done in this process is:
A spherical shell of radius \(R\) has charge \(Q\) uniformly distributed over its surface. The work done in this process is:
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Electrostatic energy of a charged spherical shell is always \(W = \frac{1}{2}QV\).
Updated On: Jun 19, 2026
- \(\frac{Q^2}{4\pi \varepsilon_0 R}\)
- \(\frac{Q^2}{8\pi \varepsilon_0 R}\)
- \(\frac{Q^2}{16\pi \varepsilon_0 R}\)
- \(\frac{Q^2}{32\pi \varepsilon_0 R}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understand the physical concept.
Work done in assembling charge on a spherical shell is equal to the electrostatic potential energy stored in the system.
Step 2: Potential on a spherical shell.
For a conducting spherical shell: \[ V = \frac{1}{4\pi \varepsilon_0}\frac{Q}{R} \]
Step 3: Work done expression.
Work required to assemble charge: \[ W = \frac{1}{2} QV \]
Step 4: Substitute value of potential.
\[ W = \frac{1}{2} Q \cdot \frac{1}{4\pi \varepsilon_0}\frac{Q}{R} \]
Step 5: Simplify expression.
\[ W = \frac{Q^2}{8\pi \varepsilon_0 R} \]
Step 6: Final conclusion.
Thus, the work done is: \[ \boxed{\frac{Q^2}{8\pi \varepsilon_0 R}} \]
Work done in assembling charge on a spherical shell is equal to the electrostatic potential energy stored in the system.
Step 2: Potential on a spherical shell.
For a conducting spherical shell: \[ V = \frac{1}{4\pi \varepsilon_0}\frac{Q}{R} \]
Step 3: Work done expression.
Work required to assemble charge: \[ W = \frac{1}{2} QV \]
Step 4: Substitute value of potential.
\[ W = \frac{1}{2} Q \cdot \frac{1}{4\pi \varepsilon_0}\frac{Q}{R} \]
Step 5: Simplify expression.
\[ W = \frac{Q^2}{8\pi \varepsilon_0 R} \]
Step 6: Final conclusion.
Thus, the work done is: \[ \boxed{\frac{Q^2}{8\pi \varepsilon_0 R}} \]
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