Question

Three concentric charged metallic spherical sheets \(A, B\) and \(C\) have radii \(a, b, c\) potentials \(V_A\), \(\mathrm{V}_{\mathrm{B}}, \mathrm{V}_{\mathrm{C}}\) and charge densities \(+\sigma,-\sigma\) and \(+\sigma\) respectively. The value of potential \(\mathrm{V}_{\mathrm{A}}\) is ( \(\varepsilon_0=\) permittivity of free space)

• A. \(\frac{\sigma}{\epsilon _{0}}(\mathrm{a}+\mathrm{b}+\mathrm{c})\)
• B. \(\frac{\sigma}{\epsilon _{0}}(-\mathrm{a}+\mathrm{b}-\mathrm{c})\)
• C. \(\frac{\sigma}{\epsilon _{0}}(\mathrm{a}-\mathrm{b}+\mathrm{c})\)
• D. \(\frac{\sigma}{3\epsilon _{0}}(\mathrm{a}+\mathrm{b}+\mathrm{c})\)

Hint:

Add the potentials due to each shell at the position of the innermost shell, using charge as surface density times area.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
For a charged spherical shell, the potential inside or on it equals \( \dfrac{Q}{4\pi\varepsilon_0 R} \) of that shell, where \( R \) is the shell radius.

Step 2: Charges on the Shells
\( Q_A=4\pi a^2\sigma \), \( Q_B=-4\pi b^2\sigma \), \( Q_C=4\pi c^2\sigma \)

Step 3: Potential at A
A is inside B and C, so it sees their constant surface potentials:
\( V_A=\dfrac{Q_A}{4\pi\varepsilon_0 a}+\dfrac{Q_B}{4\pi\varepsilon_0 b}+\dfrac{Q_C}{4\pi\varepsilon_0 c} \)

Step 4: Simplify
\( V_A=\dfrac{\sigma}{\varepsilon_0}(a-b+c) \), which is option (C).

Final Answer:
The potential of the innermost shell is sigma over epsilon-zero times (a minus b plus c). \[ \boxed{\text{(C) }\dfrac{\sigma}{\epsilon_{0}}(\mathrm{a}-\mathrm{b}+\mathrm{c})} \]