A parallel-plate capacitor has a plate area of 0.2 m 2 and a plate separation of

Question

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 2.0E6 V/m between the plates, the magnitude of the charge on each plate should be:

A) 3.5E-6 C

B) 7.1E-6 C

C) 1.4E-5 C

D) 1.8E-5 C

E) 8.9E-5 C

A parallel-plate capacitor has a plate area of 0.2 m2 and a plate separation of 0.1 mm. If the charge on each plate has a magnitude of 4.0E–6 C the potential difference across the plates is approximately:

A) 0 V

B) 4E–2 V

C) 2E2 V

D) 2E5 V

E) 4E8 V

Explanation / Answer

1) A) 3.5E-6 C

E = Q/(A*epsilon)

Q = E*A*epsilon

= 2*10^6*0.2*8.854*10^-12

= 3.5*10^-6 C

2) C) 2E2 V

C = A*epsilon/d

= 0.2*8.854*10^-12/(0.1*10^-3)

= 1.77*10^-8 C

V = Q/C

= 4*10^-6/(1.77*10^-8)

= 226 volts

= 2E2 V

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