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?

(a) surface area of the plate will be equal to

(b) Charge on the capacitor is equal to

Explanation:

We have given spacing between the plates d = 0.05 mm =

Value of capacitance

(A) Capacitance of a parallel plate capacitor is equal to

So

So surface area of the plate will be equal to

(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

So

b mirrors submit a iisner? e. for which optical devices does d sometimes have a positive value? o a a and b o b. a and c o c. b and d 0 d. b and cb

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explanation: