Question

A charge \(Q\) is distributed over two concentric hollow spheres of radii \(a\) and \(b\) \((a>b)\), so that the surface charge densities are equal. The potential at the common centre is \(\frac{1}{4\pi \varepsilon_0}\) times

• A. \(Q \left(\frac{a+b}{a^2 + b^2}\right)\)
• B. \(2Q \left(\frac{a+b}{a^2 + b^2}\right)\)
• C. \(Q\)
• D. \(\frac{Q}{2} \left(\frac{a+b}{a^2 + b^2}\right)\)
• E. \(\frac{Q}{4} \left(\frac{a+b}{a^2 + b^2}\right)\)

Hint:

Potential inside a spherical shell is same as on surface.

Answer 1

The Correct Option is A

Concept: Potential due to spherical shell: \[ V = \frac{1}{4\pi \varepsilon_0} \frac{Q}{R} \] Surface charge density: \[ \sigma = \frac{Q}{4\pi R^2} \]

Step 1: Equal surface charge densities. \[ \frac{Q_1}{4\pi a^2} = \frac{Q_2}{4\pi b^2} \Rightarrow \frac{Q_1}{a^2} = \frac{Q_2}{b^2} \]

Step 2: Express charges. \[ Q_1 = \frac{a^2}{b^2} Q_2 \]

Step 3: Total charge. \[ Q = Q_1 + Q_2 = \frac{a^2}{b^2}Q_2 + Q_2 = Q_2 \left(\frac{a^2 + b^2}{b^2}\right) \]

Step 4: Find \(Q_2\). \[ Q_2 = \frac{Q b^2}{a^2 + b^2} \quad,\quad Q_1 = \frac{Q a^2}{a^2 + b^2} \]

Step 5: Potential at centre. \[ V = \frac{1}{4\pi\varepsilon_0} \left(\frac{Q_1}{a} + \frac{Q_2}{b}\right) \]

Step 6: Substitute values. \[ V = \frac{1}{4\pi\varepsilon_0} \left(\frac{Qa^2}{a(a^2+b^2)} + \frac{Qb^2}{b(a^2+b^2)}\right) \]

Step 7: Simplify. \[ = \frac{1}{4\pi\varepsilon_0} \cdot \frac{Q(a+b)}{a^2 + b^2} \]

Step 8: Conclusion. \[ \boxed{Q\left(\frac{a+b}{a^2 + b^2}\right)} \]