A parallel plane condenser is filled with two dielectrics as shown in figure. Area of each plate is A in m2 and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance in farad will be:

Question

A parallel plane condenser is filled with two dielectrics as shown in figure. Area of each plate is A in m2 and the separation is d metre. The dielectric constants are K1 and K2 respectively. Its capacitance in farad will be: (A)  (2ε 0A/d)( (K1+K2/K1K2) )  (B) (2 ∈0 A / d )(( K 1 K 2/ K 1+ K 2)) (C)  (ε 0A/d)( (K1+K2/2K1K2) )  (D)  (ε 0AK1+K2/2(d2K1+d1K2))

Tardigrade Admin · Accepted Answer

From the formula C=(ε 0A/(1)K1+(1)K2=(ε 0A/t)( (K1K2/K1+K2) ) =(2ε 0A/d)( (K1K2/K1+K2) )

Q. A parallel plane condenser is filled with two dielectrics as shown in figure. Area of each plate is A in m2 and the separation is d metre. The dielectric constants are $K_{1}$ and $K_{2}$ respectively. Its capacitance in farad will be:

$\frac{2 ε _{0} A}{d} (\frac{K _{1} + K _{2}}{K _{1} K _{2}})$

$\frac{2 \in _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

$\frac{ε _{0} A}{d} (\frac{K _{1} + K _{2}}{2 K _{1} K _{2}})$

$\frac{ε _{0} A K _{1} + K _{2}}{2 ( d _{2} K _{1} + d _{1} K _{2} )}$

From the formula $C = \frac{ε _{0} A}{\frac{1}{K _{1}} + \frac{1}{K _{2}}} = \frac{ε _{0} A}{t} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$ $= \frac{2 ε _{0} A}{d} (\frac{K _{1} K _{2}}{K _{1} + K _{2}})$

Q. A parallel plane condenser is filled with two dielectrics as shown in figure. Area of each plate is A in m2 and the separation is d metre. The dielectric constants are K1​ and K2​ respectively. Its capacitance in farad will be:

d2ε0​A​(K1​K2​K1​+K2​​)

d2∈0​A​(K1​+K2​K1​K2​​)

dε0​A​(2K1​K2​K1​+K2​​)

2(d2​K1​+d1​K2​)ε0​AK1​+K2​​