A water-filled parallel plate capacitor has a plate area of 2.7 cm2 and plate se

Question

A water-filled parallel plate capacitor has a plate area of 2.7 cm2 and plate separation of 1.5 mm. The potential difference between its plates is held at 6.7 V. Calculate the magnitude of the electric field between its plates, the charge stored on each plate, and the charge stored on each plate after water is replaced by air. (a) the magnitude of the electric field between its plates V/m (b) the charge stored on each plate nC (c) the charge stored on each plate after water is replaced by air pC

Explanation / Answer

E=v/d

Putting values

E=4466.7 N/C

Capacitor=k*epsilon*area/d=1.28*10^(-10) farad

Charge=capacitor*voltage=8.57*10^(-10) C

If water is replaced by air

Charge=(C/80)*6.7=1.07*10^(-11)

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