First slide

Capacitance

Question

A $2 μ F$ parallel capacitor is connected to a 20 volt battery as shown in figure. Now the plates of the capacitor are pulled apart to increase their separation to two times its present value in 0.5 second. Then average current flown through the battery during this period is _______

Moderate

A

$20 μ A$
B

$40 μ A$
C

0.2 mA
D

$4 μ A$

Solution

$\begin{array}{l} I_{a v g} = \frac{|Δ Q|}{Δ t} = \frac{C V - \frac{C}{2} . V}{Δ t} = \frac{\frac{1}{2} (C V)}{Δ t} \\ = \frac{\frac{1}{2} \times 2 \times 20}{0.5} μ A = 40 μ A \end{array}$

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