Question

Find the capacitance of a system of three parallel plates each of area A separated by distances d1 and d2. The space between them is filled with dielectrics of relative permittivities E1 and E2. The dielectrics of free space is Eo

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
$\frac{1}{c(eq)} = \frac{1}{c(1)} + \frac{1}{c(2)}$

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

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

Answer 2

Answer:

C(eq) = £° A (1 /( (d1/£1)+ (d2/£2))

Explanation:

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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