Discussion
A conducting sphere of radius R has charge +Q on its surface. If the charge on the surface is doubled and its radius is halved, the energy associated with the electric field will :
Answer: C
The energy will increase eight times.
The energy stored in the field of a charged conducting sphere is:
\[ \begin{aligned} U &= \frac{Q^{2}}{8\pi\varepsilon_{0}R} \\ U' &= \frac{(2Q)^{2}}{8\pi\varepsilon_{0}(R/2)} = 8U \end{aligned} \]
Doubling the charge multiplies the energy by 4, and halving the radius doubles it again. 'Four times' accounts only for the charge.
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