A capacitor of capacitance $1 μ F$ is filled with two dielectrics of dielectric constants 4 and 6. What is the new capacitance?

$10 μ F$

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$5 μ F$

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$4 μ F$

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$7 μ F$

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Solution

The correct option is B
$5 μ F$

From the above problem,

$C = \frac{ϵ_{0} A}{2 d} (K_{1} + K_{2})$

$= \frac{C_{0}}{2} (K_{1} + K_{2})$

$= \frac{1}{2} (4 + 6) = 5 μ F$

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A capacitor of capacitance 1 μF is filled with two dielectrics of dielectric constants 4 and 6. What is the new capacitance?

The correct option is B 5 μF From the above problem, C=ϵ0A2d(K1+K2) =C02(K1+K2) =12(4+6)=5 μF

A capacitor of capacitance $1 μ F$ is filled with two dielectrics of dielectric constants 4 and 6. What is the new capacitance?

The correct option is B
$5 μ F$

From the above problem,

$C = \frac{ϵ_{0} A}{2 d} (K_{1} + K_{2})$

$= \frac{C_{0}}{2} (K_{1} + K_{2})$

$= \frac{1}{2} (4 + 6) = 5 μ F$