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Question

Physics Question on electrostatic potential and capacitance

What is the area of the plates of a 3 F3\, F parallel plate capacitor, if the separation between the plates is 5 mm5\, mm ?

A

9.281×109m29.281 \times 10^{9} m ^{2}

B

4.529×109m24.529 \times 10^{9} m ^{2}

C

1.694×109m21.694 \times 10^{9} m ^{2}

D

12.981×109m212.981 \times 10^{9} m ^{2}

Answer

1.694×109m21.694 \times 10^{9} m ^{2}

Explanation

Solution

Let the area be A The separation between the plates d=5 mm=0.005 md = 5\, mm = 0.005\, m We know capacitance C=ε0AdC=\frac{\varepsilon_{0}A}{d} ⇒A=Cdε0\Rightarrow A=\frac{Cd}{\varepsilon_{0}} =4π×3×0.005×9×109= 4\pi \times 3 \times 0.005 \times 9 \times 10^{9} =1.69×109 m2=1.69 \times10^{9}\, m^{2}