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.
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