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Electrostatic Potential And Capacitance

Question

What is the area of the plates of a parallel plate capacitor of capacitance of 2F with separation between the plates 0.5 cm?

Solution

Given, a parallel plate capacitor.
Capacitance of plates, C= 2F
Seperation between the plates, d=0.5 cm= $0.5 \times 10^{- 2} m$

Capacitance of a parallel plate capacitor is given by the formula C= $\frac{ε_{\circ} A}{d}$

Therefore, area of the plates, A = $\frac{Cd}{ε_{\circ}} = \frac{2 \times 0.5 \times 10^{- 2}}{8.854 \times 10^{- 12}} = 11.3 \times 10^{10} m^{2}$

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Electrostatic Potential And Capacitance

What is the area of the plates of a parallel plate capacitor of capacitance of 2F with separation between the plates 0.5 cm?

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