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

Question

A parallel-plate capacitor with plate area A, plate separation d and a capacitance of C_0 =15 mu F in the air. Now a dielectric medium with dielectric constant x = 2.85 is inserted into the plates. First, the medium occupies all the space constant between the plates (see Figure a), and the capacitance is C_1. In the second case, it occupies upper half space between the two plates (see Figure b). The capacitance is C_2. Express the capacitance of a parallel-plate capacitor in the air in terms of epsilon_0, A and d. C_0 = Express the capacitance of a parallel-plate capacitor with a dielectric inside in terms of x, epsilon _0, A and d. Express C_l in terms of C_0. Calculate the numerical value of C_1 in mu F. In figure B, express the capacitance of the upper half of the capacitor, C_u, in terms of x, epsilon_0, A, and d. In figure B, express the capacitance of the lower half of the capacitor, C_d, in terms of epsilon_0, A and d. How are the upper half and lower half connected to each other, in series or in parallel? Express the total capacitance C_2 in terms of x, epsilon_0, A, and d. Express the total capacitance C_2 in terms of C_0 Calculate the numerical value of C_2 in mu F.

Explanation / Answer

a)The capacitance of a parallel plate capacitor is related to Area, distance and permittivity as:

C = epsilon0 A/d

b)when a dielectric with constant "x" is inserted the capacitance becomes:

C = x epsilon0 A/d

c)After inserting the dielectric with constant "x", the capacitance now becomes:

C1 = x C0

d)Given,

C0 = 15 uF = 15 x 10^-6 F and x = 2.85

C1 = x C1 = 2.85 x 15 x 10^-6 = 42.75 x 10^-6 F

Hence, C1 = 42.75 uF

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