Question

A capacitor consists of three parallel plates of equal area A.The distance between them is D1 and D2. Dielectric material having permittivity e1 and e2 is present between the plates. express this capacitance is terms of k1 and k2​

Answer 1

Final Answer :

C(eq) = £° A (1 /( (d1/£1)+ (d2/£2))

Steps and Understanding :

1) Relative Permitivity/ Dielectric Constant :

£1,£2.

Capacitance of a two parallel plates separated by d1 distance, filled by dielectric of dielectric Constant £1 is

C (1) = £1£°A / (d1 ).

2) Similarly,

C(2) = £2£°A/(d2)

3) Given us system of two series plate,

so

=> 1/c(eq) = d1 / (£1£° A) + d2 / (£2£°A)

=> C(eq) = £° A ( 1/ (d1/£1) + 1/(d2/£2))

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Answer 2

$C=\frac{Ae_1e_2k_1k_2}{D_1e_1k_1+D_2e_2K_2}$

Explanation:

Capacitance of a two parallel plates separated by D1 distance, filled by dielectric of dielectric Constant $k_1$ is

$C_1=\frac{Ae_1k_1}{D_1}$

similarly for second,

$C_2=\frac{Ae_2k_2}{D_2}$

There are two capacitors C  _1  and C  _2  and they are in series.

So equivalent capacitance is

$\frac{1}{C} =\frac{1}{C_1} +\frac{1}{C_2}$

Substitute the value, we get

$C=\frac{Ae_1e_2k_1k_2}{D_1e_1k_1+D_2e_2K_2}$

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