question_answer

A spherical rubber balloon carries a charge that is uniformly distributed over its surface. As the balloon is blown up; how does E vary for points (i) inside the balloon, (ii) on the surface of the balloon and (iii) outside the balloon?

Answer:

                (i) For points inside the balloon, \[E=0\] (ii) As the balloon is blown up, surface charge density \[\sigma \] decreases and so the field, \[(E=\sigma /{{\varepsilon }_{0}})\]on its surface decreases. (iii) For points outside the balloon, \[E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{r}^{2}}}\] As the balloon is blown up, the charge enclosed by the Gaussian surface remains same, so E does not change.