Problem 5 please A variable parallel plate capacitor consists of two metal plate

Question

Problem 5 please

A variable parallel plate capacitor consists of two metal plates with area A = 10-2 m2, separated from one another by a spring with unscratched length d0=1.0mm, and stiffness constant k = A times 10-4 N/m, as shown in the figure. What is the electric field E between the plates when the capacitor is given a charge of 0.1 nC? Considering that the sum of the energy stored in the elactric field, 1/2epsilon0E2Ad, and the energy stored in the spring, 1/2k (d - d0)2 , should reach a minimum in equilibrium, find the plate seperation d for a charge of 0.1 nC.

Explanation / Answer

a)

E=q/A*e0

=0.1*10^-9/(10^-2*8.854*10^-12)

=1129.433 N/C or V/m

b)

total energy Unet=U1+U2 should reach minimum if derivative of the U should equals to zero

(1/2eoE^2*A)+1/2K*2*(d-d0)=0

(eoE^2*A)+K*2*(d-d0)=0

d=d0-(eoE^2*A)/2k

d=(1*10^-3) - (8.85*10^-12*(1129.43)^2*10^-2/2*10^-4)

d=0.436 mm

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