A parallel capacitor has a capacitance of 1.31 nF. There is a charge of magnitud

Question

A parallel capacitor has a capacitance of 1.31 nF. There is a charge of magnitude 0.792 micro charge on each plate. (A) what is the potential difference between the two plates ? (B) if the plate seperation increases by a factor of 3, while the charge is kept constant, what will happen to the potential difference ? A parallel capacitor has a capacitance of 1.31 nF. There is a charge of magnitude 0.792 micro charge on each plate. (A) what is the potential difference between the two plates ? (B) if the plate seperation increases by a factor of 3, while the charge is kept constant, what will happen to the potential difference ?

Explanation / Answer

Q = CV
V =Q/C
V = 0.792 x10^-6 / 1.31 x10^-9
V = 0.604 x10^3
V= 604.58 V

if the plate seperation increases by a factor of 3,
the potential difference increases by 3 times

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