Question

part b
the capacitor is now disconnected from the battery, and the plates of the capacitor are then slowly pulled apart until the separation reaches 3d. find the new energy (u_1) of the capacitor after this process.
express your answer in terms of (a), (d), (v), and (epsilon_0).
view available hint(s)
(u_1 =)

Explanation:

Step1: Recall capacitance formula

The capacitance of a parallel - plate capacitor is $C=\frac{\epsilon_0A}{d}$. Initially, let the capacitance be $C_0 = \frac{\epsilon_0A}{d}$. When the plate separation is changed to $3d$, the new capacitance $C_1=\frac{\epsilon_0A}{3d}$.

Step2: Recall charge conservation

Since the capacitor is disconnected from the battery, the charge $Q$ is conserved. Initially, $Q = C_0V=\frac{\epsilon_0A}{d}V$. After the change, $Q = C_1V_1$, where $V_1$ is the new potential difference. Since $Q$ is constant, $\frac{\epsilon_0A}{d}V=\frac{\epsilon_0A}{3d}V_1$, and we can solve for $V_1 = 3V$.

Step3: Recall energy formula for capacitor

The energy stored in a capacitor is $U=\frac{1}{2}CV^{2}$. For the new capacitor, $U_1=\frac{1}{2}C_1V_1^{2}$. Substitute $C_1=\frac{\epsilon_0A}{3d}$ and $V_1 = 3V$ into the energy formula:
\[

$$\begin{align*} U_1&=\frac{1}{2}\times\frac{\epsilon_0A}{3d}\times(3V)^{2}\\ &=\frac{1}{2}\times\frac{\epsilon_0A}{3d}\times9V^{2}\\ &=\frac{3\epsilon_0AV^{2}}{2d} \end{align*}$$

\]

Answer:

$\frac{3\epsilon_0AV^{2}}{2d}$