Question

A spherical balloon of radius $a$ is charged. The energy density in the electric field at point $P$ shown in the figure given below is $W$. If the balloon is inflated to a radius $b$ without altering its charge, what is the energy density at $P$?

• (A) $W{\left(\frac{b}{a}\right)}^3$
• (B) $W{\left(\frac{b}{a}\right)}^2$
• (C) $W\left(\frac{b}{a}\right)$
• (D) $W$

Answer 1

The Correct Option is D

Explanation:
The energy density at $P$ is $W$.Explanations is given as:Gauss Law states that the net electric flux through a closed surface enclosing a volume equals the charge enclosed by it.${\oint }_s\overrightarrow{E}\cdot \overrightarrow{ds}=\frac{Q}{{\epsilon }_o}$Since $\overline{D}={\epsilon }_o\overrightarrow{E}$, we can write:${\oint }_s\overline{D}\cdot \overrightarrow{ds}=Q$$\overline{D}$ is the electric flux density.Observations:1) For a Gaussian sphere around point $P$, the net charge enclosed by it will not change with a change in radius.2) Since the net outward electric field through any closed surface is equal to the charge enclosed by it, the electric field outside the spherical balloon will not change with the change in its radius.3) So, the energy density at point $P$ will be the same as $W$ for the inflated radius $b$ of the balloon.Hence, the correct option is (D).