A certain capacitor has a capacitance of 5.0 $\mu$F. After it is charged to 5 $\mu$C and isolated, the plates are brought closer together so its capacitance becomes 10 $\mu$F. The work done by the agent is about: A) 0 B) 1.25 *10^-⁶ J C) -1.25 * 10^-⁶ J D) 8.3* 10^-⁷ J E) -8.3 * 10^-⁷ J

Question

Quizplus · Accepted Answer

The work done by the agent is given by $W=(1/2)(C_2V_2^2-C_1V_1^2)$, where $C_1$ and $V_1$ are the original capacitance and charge, and $C_2$ and $V_2$ are the final capacitance and charge. 
Plugging in the given values, we get $W=(1/2)(10\mu F)(0-5^2) - (1/2)(5\mu F)(5^2-5^2) = -12.5$ $\mu J$. Since the work done is negative, the agent has actually extracted energy from the system, so the answer should be negative. The closest answer choice to this is C, which gives the value $-1.25$ $\mu J$, which is the answer to the nearest tenth of a microjoule.

A Certain Capacitor Has a Capacitance of 5 μ\muμ F After It Is Charged to 5

A Certain Capacitor Has a Capacitance of 5 $\mu$ F After It Is Charged to 5