The two plates of a parallel plate capacitor have are charged to plus minus 4.0n

Question

The two plates of a parallel plate capacitor have are charged to plus minus 4.0nC when the voltage drop across the capacitor is 2.0 V. What is the capacitance? 8.0 nF 0.50 nF 0.125 nF 2.0 nF The electric field in a parallel plate capacitor is 5000 V/m. If the separation of the plates is 2.0 mm, what is the voltage difference between the two plates? 10 V 10,000 V 2,500 V 4.0 times 10^-4 V 2.5 times 10^-6 V What is the magnitude of the magnetic field due to a long straight wire carrying a current of

Explanation / Answer

part 6 )

C = Q/V

C = 4 * 10^-9 / 2 = 2 x 10^-9 F = 2 nF

part 7 )

V = E.d

d = 2 mm = 2 x 10^-3 m

V = 5000 x 2 x 10^-3

V = 10 V

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