Question
A parallel-plate capacitor has plate area $25.0cm^2$ and a separation of 2.00mm between the plates. The capacitor is connected to a battery of 12.0V:
  1. Find the charge on the capacitor.
  2. The plate separation is decreased to 1.00mm. Find the extra charge given by the battery to the positive plate.

Answer



Plate area $A = 25cm^2 = 2.5 \times 10^{-3}m$
Separation $d = 2mm = 2 \times 10^{-3}m$
Potential $v = 12v$​​​​​​​
  1. We know
$\text{C}=\frac{\in_0\text{A}}{\text{d}}$
$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{2\times10^{-3}}$
$=11.06\times10^{-12}\text{F}$
$\text{C}=\frac{\text{q}}{\text{v}}$
$\Rightarrow11.06\times10^{-12}=\frac{\text{q}}{12}$
$\Rightarrow\text{q}_1=1.32\times10^{-10}\text{C}.$
  1. Then d = decreased to 1mm
$\therefore\ \text{d}=1\text{mm}=1\times10^{-3}\text{m}$
$\text{C}=\frac{\in_0\text{A}}{\text{d}}=\frac{\text{q}}{\text{v}}$
$=\frac{8.85\times10^{-12}\times2.5\times10^{-3}}{1\times10^{-3}}=\frac{2}{12}$
$\Rightarrow\text{q}_2=8.85\times2.5\times12\times10^{-12}$
$\Rightarrow\text{q}_2=2.65\times10^{-10}\text{C}.$
$\therefore$ The extra charge given to plate $= (2.65 - 1.32) \times 10^{-10} = 1.33 \times 10^{-10}C.$

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