Question

A spherical charged conductor has $σ$ as the surface density of charge. The electric field on its surface is E. If the radius of the sphere is doubled keeping the surface density of charge unchanged, what will be the electric field on the surface of the new sphere ?

Easy

Solution

$\begin{array}{l} E = \frac{1}{4 {πε}_{0}} \cdot \frac{Q}{R^{2}} \\ = \frac{1}{4 {πε}_{0}} \times \frac{4 {πR}^{2} σ}{R^{2}} = \frac{σ}{ε_{0}} \end{array}$
This is independent of radius and depends on $σ$ . Hence, the electric field on the surface of the new sphere will be E.

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