A parallel plate capacitor has a capacitance of 1.20 nF. There is a charge of ma

Question

A parallel plate capacitor has a capacitance of 1.20 nF. There is a charge of magnitude 0.800 C on each plate. (a) What is the potential difference between the plates? (b) If the plate seperation is doubled,while the charge is kept constant, what will happen to the potential difference?  

Explanation / Answer

We know that Q = CV Where    C = o A / d ==>    Q / V = o A /d ==>    V = Q d /o A                 (a)   V = Q / C = 0.8 x10-6 / 1.2 x 10-9                           = 666.66 V (b) V = Q d / o A         If the plate separationis doubled,while the charge is kept constant, thepotential difference also doubled.

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