Problem: Given a parallel-plate capacitor with plates of area A separated by a distance l. Consider the quantities, (i) capacitance C, (ii) magnitude E of the electric field between the plates, (iii) magnitude of the charge Q on the plates, (iv) potential difference ΔV between the plates, and (v) energy U stored in the capacitorAssume we apply a given potential difference ΔV0 to the plates. Suppose we double the plate separation while keeping the potential difference constant, calculate the quantities (i) - (v), for this case, and express each in terms of the original value, e.g., C = 2C0, E = (1/4)E0, etc. (Obviously, ΔV = ΔV0).

FREE Expert Solution

The capacitance of a parallel plate capacitor:

C=ε0Ad=0Ad

The charge stored on a capacitor:

Q=CV

Energy stored by a capacitor:

U=12CV2=12QV=Q22C

The electric field in a parallel plate capacitor:

E=Qε0A=Vd

(i)

Using the expression for the capacitance:

C0 = ε0A/l

C = ε0A/2l

C/C0 = (ε0A/2l)/(ε0A/l) = 1/2

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Problem Details

Given a parallel-plate capacitor with plates of area A separated by a distance l. Consider the quantities, (i) capacitance C, (ii) magnitude E of the electric field between the plates, (iii) magnitude of the charge Q on the plates, (iv) potential difference ΔV between the plates, and (v) energy U stored in the capacitor

Assume we apply a given potential difference ΔV0 to the plates. Suppose we double the plate separation while keeping the potential difference constant, calculate the quantities (i) - (v), for this case, and express each in terms of the original value, e.g., C = 2C0, E = (1/4)E0, etc. (Obviously, ΔV = ΔV0).

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