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Capacitors — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsCapacitors5 Marks
Question
A parallel-plate capacitor has plate area 25.0cm2 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 = 25cm2 = 2.5 × 10-3m

Separation d = 2mm = 2 × 10-3m

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) × 10-10 = 1.33 × 10-10C.

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