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# An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.40 ...

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.40 mm. If a 18.4-V potential difference is applied to these plates, calculate the following. (a) the electric field between the plates  b. the capacitance in  pF   (c) the charge on each plate  in  pC

To determine: a) The electric field between the plates E

b) The capacitance C

c) The charge on each plate Q

Given data: Surface are of plate of parallel plate capacitor As = 7.60 cm2 = 7.6×10-4 m2

Gap between the two plates d = 1.40 mm = 1.4×10-3 m

Potential difference is applied to these plates V = 18.4 V

a) The electric field E between the plates of a parallel plate capacitor is given by  b) Q=CV But c) Charge on each plate Q is given by

Q=CV Note that each plate of the parallel plate capacitor will have charge of magnitude 88.4377pC. One plate will have +88.4377p and another will have -88.4377pC charge. Similar Questions

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