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Amanda [17]
3 years ago
9

The spacing between the plates of a 1.0 μF capacitor is 0.050 mm. a) What is the surface area of the plates? b) How much charge

is on the plates if this capacitor is charged to 1.5 V?
Physics
1 answer:
nataly862011 [7]3 years ago
5 0

Answer:

(a) surface area of the plate will be equal to 1.129m^2

(b) Charge on the capacitor is equal to 1.5\times 10^{-6}C

Explanation:

We have given spacing between the plates d = 0.05 mm = 5\times 10^{-5}m

Value of capacitance C=1\mu F=10^{-6}F

(A) Capacitance of a parallel plate capacitor is equal to C=\frac{\epsilon _0A}{d}

So 10^{-6}=\frac{8.85\times 10^{-12}\times A}{10^{-5}}

A=1.129m^2

So surface area of the plate will be equal to 1.129m^2

(B) It is given that capacitor is charged by 1.5 volt

So voltage V = 1.5 volt

Charge on the capacitor is equal to Q=CV

So Q=1.5\times 10^{-6}C

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