# A If a capacitor has opposite 4.9 \mu C charges on the plates, and an electric field of 2.3 kV/mm...

## Question:

A If a capacitor has opposite {eq}4.9 \mu C {/eq} charges on the plates, and an electric field of {eq}2.3 kV/mm {/eq} is desired between the plates, what must each plate's area be?

## Capacitance:

The electric field between the capacitor is given by the ratio of the potential difference to the distance between the plates. The electric field between the plate remains constant while the potential difference changes.

## Answer and Explanation:

**Given data**

Charge on the plates {eq}Q = 4.9 \ \mu C {/eq}

Electric field {eq}E = 2.3 \ kV/mm \\ E = 2.3 \times 10^{6} \ V/m {/eq}

Now, the charge on the capacitor is given by

{eq}Q = C\times V \\ Q = \dfrac{\epsilon_{o}A}{d} \times (Ed) \\ Q = \epsilon_{o}A \times E \\ (4.9 \times 10^{-6} \ C) = (8.85 \times 10^{-12}) \times A \times (2.3\times 10^{6} \ V/m) \\ A = 0.241 \ m^{2} {/eq}

where

- A is the area of the plates
- {eq}(C) =\dfrac{\epsilon_{o}A}{d} {/eq} is the capacitance

- E is the electric field
- d is the distance between the plates

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