First slide

Capacitance

Question

A potential difference of 2000 V is established between the parallel plates of a capacitor with air between the plates. If the air becomes conducting when the electric field exceeds 3 x 10⁶ N/C, the minimum separation between the plates should be

Moderate

A

0 .01m
B

0.001m
C

0. 667 mm
D

1.34 mm

Solution

Let the minimum separation should be x. Then
$\begin{array}{l} E = \frac{V}{d} = \frac{2000}{x} = 3 \times 10^{6} \\ ∴ x = \frac{2000}{3 \times 10^{6}} = \frac{2}{3} \times 10^{- 3} = 0 .667 mm \end{array}$

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