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vampirchik [111]
3 years ago
8

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
Physics
1 answer:
Effectus [21]3 years ago
8 0
<span>1.8 × 10-6 C </span> 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 
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3 years ago
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