The charged balloon in FIGURE Q24.7 expands as it is blown up, increasing in size from the initial to final diameters shown. Do the electric field strengths at points 1, 2, and 3 increase, decrease, or stay the same? Explain your reasoning for each.

Given Figure :

Theory used :

At all sites in electrostatic equilibrium inside a conductor, the Electric Field is zero. However, all surplus charges on the conductor accumulate on the outside surface, and as further charges are added, they spread out on the outer surface until they reach the electrostatic equilibrium points. The electric field at the surface of a charged conductor in the form

$E_{surface}=\frac{\eta }{{\epsilon }_0}$, where $\eta$ is the surface charge density.

The surface charge density is not constant and is a physical parameter that relies on the conductor's geometry. Calculating the surface charge density using the electric field outside the conductor :

Because the electric field at point 1 is the same at both the beginning and the end of the balloon, the electric field at point 1 is the same.

The electric field at point 2 is weaker. Because point 2 is outside the balloon at first, an electric field exists due to charges on the surface; however, as the balloon expands, point 2 becomes inside the balloon, and the electric field vanishes. The electric field shrinks in this situation.

The electric field at point 3 is same. Point 3 lies outside the balloon both before and after the expansion, therefore the electric field is the same.