First slide

Electrostatic potential

Question

Concentric metallic hollow spheres of radii R and 4R hold charges $Q_{1}$ and $Q_{2}$ respectively. Given that surface charge densities of the concentric spheres are equal, the potential difference V(R) – V(4R) is :

Moderate

A

$\frac{3 Q_{1}}{4 π ε_{0} R}$
B

$\frac{Q_{2}}{4 π ε_{0} R}$
C

$\frac{3 Q_{1}}{16 π ε_{0} R}$
D

$\frac{3 Q_{2}}{4 π ε_{0} R}$

Solution

Surface charge densities are equal
   $∴ \frac{Q_{1}}{4 π R^{2}} = \frac{Q_{2}}{4 π {(4 R)}^{2}}$
   $\Rightarrow Q_{2} = 16 Q_{1}$
Potential of inner shell $= \frac{k Q_{1}}{R} + \frac{k Q_{2}}{4 R}$
Potential of outer shell =    $\frac{k Q_{1}}{4 R} + \frac{k Q_{2}}{4 R}$
$∴$ Potential difference between shells = $\frac{3 k Q_{1}}{4 R}$ = $\frac{3 Q_{1}}{16 π ε_{0} R}$

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