A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm. to obtain an electric field of 2.0 * 106 v/m between the plates, the magnitude of the charge on each plate should be

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Maren Fitzgerald

Physics

13 December, 01:16

A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm. to obtain an electric field of 2.0 * 106 v/m between the plates, the magnitude of the charge on each plate should be

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Santos Reynolds · Answer 1

1.8 * 10-6 C You need to know that charge Q=C*V, that the capacitance C=epsilon_0*A/d and that the electric field between plates is given by E=V/d. Now use these three equations to solve for Q, divide by 2 and get answer