A parallel-plate capacitor with plate area A , plate separation d and a capacita

Question

A parallel-plate capacitor with plate area A, plate separation d and a capacitance of C0 = 16 ?F in the air. Now a dielectric with dielectric constant ? = 3 is inserted. In the first case, it occupies all space between two plates (see Figure a). The capacitance is C1. In the second case, it occupies upper half space between two plates (see Figure b). The capacitance is C2.

Part (a): Calculate the numerical value of C1 in ?F.

Part (b): In figure B, express the capacitance of the upper half between the two plate Cu in terms of ?, ?0, A, and d.

Part (c): In figure B, express the capacitance of the lower half between the two plate Cd in terms of ?0, A and d.

Part (d): How are the upper half and lower half connected to each other, in series or in parallel?

Part (e): Express the total capacitance C2 in terms of ?, ?0, A, and d.

Part (f): Express the total capacitance C2 in terms of C0.

Part (g): Calculate the numerical value of C2 in ?F.

Explanation / Answer

a) Co = A*epsilon/d

b) C1 = k*A*epsilon/d

c) C1 = K*co

d) C1 = K*co = 48 micro F

e) Ca = K*A*epsilon/(d/2) = 2*K*Co

f) Cb = A*epsilon/(d/2) = 2*Co

g) in series

h) C2 = Ca*Cb/(Ca+Cb) = 2*k*Co/(1+k)

C2 = 2*k/(1+k)*A*epsilon/d

i)C2 = 2*k*Co/(1+k)

j) C2 = 2*k*Co/(1+k) = 24 micro F

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