First slide

Potential energy

Question

A parallel plate capacitor has plate area A and separation d. It is charged to a potential difference V_o. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plate is

Moderate

A

$\frac{3 ε_{0} {AV}_{0}^{2}}{d}$
B

$\frac{ε_{0} {AV}_{0}^{2}}{2 d}$
C

$\frac{ε_{0} {AV}_{0}^{2}}{3 d}$
D

$\frac{ε_{0} {AV}_{0}^{2}}{d}$

Solution

Here $C = \frac{ε_{0} A}{d}$ and $Q = CV = \frac{ε_{0} {AV}_{0}}{d}$
$∴$ Energy stored, $E_{1} = \frac{Q^{2}}{2 C} = \frac{ε_{0}^{2} A^{2} V_{0}^{2}}{2 d^{2} C}$
When battery is disconnected, Q remains the same. As d increases to 3 times, C decreases to 3 times.
$∴ E_{2} = \frac{3 Q^{2}}{2 C} = \frac{3 ε_{0}^{2} A^{2} V_{0}^{2}}{2 d^{2} C}$
Now, Work done = E₂ - E₁
$\begin{array}{l} ∴ W = \frac{3 ε_{0}^{2} A^{2} V_{0}^{2}}{2 d^{2} C} - \frac{ε_{0}^{2} A^{2} V_{0}^{2}}{2 d^{2} C} = \frac{ε_{0}^{2} A^{2} V_{0}^{2}}{d^{2} C} \\ = \frac{ε_{0}^{2} A^{2} V_{0}^{2}}{d^{2} \times (ε_{0} A / d)} = \frac{ε_{0} {AV}_{0}^{2}}{d} \end{array}$

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