Question #107363
an airfilled parallelplate capacitor has plates of area 2.30cm^2 seperated by 1.50 mm. a)find the value of its capacitance . the capacitor is connected to a 120 v battery. b)what is the charge on the capacitor c)what is the magnitude of the uniform electric field between the plates?
1
Expert's answer
2020-04-01T10:15:44-0400

(a) The capacitance

C=ϵ0Ad=8.85×1012×2.30×1041.50×103=1.36×1012FC=\frac{\epsilon_0A}{d}=\frac{8.85\times 10^{-12}\times 2.30\times 10^{-4}}{1.50\times 10^{-3}}=1.36\times 10^{-12}\:\rm F

(b) The charge on plates

q=CV=1.36×1012×120=1.63×1010C.q=CV=1.36\times 10^{-12}\times 120=1.63\times 10^{-10}\:\rm C.

(c) The electric field between plates

E=Vd=1201.50×103=80000V/m.E=\frac{V}{d}=\frac{120}{1.50\times 10^{-3}}=80\: 000\:\rm V/m.

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