Space between plates is filled by two dielectrics K1 - K2- 2-3rd part is filled by K1 - 1-3rd is filled by K2- find effective dielectric constant- Area of the plates is A-

Both capacitors connected parallel. So C_eff = C_1 + C_2 =ϵ_0 2A/3 K_1d + ϵ_0 A/3K_2d ( ϵ_0 A/d )K_eff = ϵ_0 A/d ( 2 K_1/3 +K_2/3 ) K_eff =2K_1 +K_2/3 ...

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Byju's Answer

Standard XII

Physics

New Capacitance in Dielectric

Space between...

Question

Space between plates is filled by two dielectrics $K_{1} & K_{2}, {\frac{2}{3}}^{r d}$ part is filled by $K_{1} & {\frac{1}{3}}^{r d}$ is filled by $K_{2}$ , find effective dielectric constant. Area of the plates is $A$ .

\frac{2 K_{1} + K_{2}}{3}

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\frac{K_{1} + 2 K_{2}}{3}

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\frac{3 K_{1} + K_{2}}{4}

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\frac{K_{1} + 3 K_{2}}{4}

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Solution

The correct option is A $\frac{2 K_{1} + K_{2}}{3}$

Both capacitors connected parallel. So

$C_{e f f} = C_{1} + C_{2}$

$= \frac{ϵ_{0} \frac{2 A}{3} K_{1}}{d} + \frac{ϵ_{0} \frac{A}{3} K_{2}}{d}$

$(\frac{ϵ_{0} A}{d}) K_{e f f} = \frac{ϵ_{0} A}{d} (\frac{2 K_{1}}{3} + \frac{K_{2}}{3})$

$K_{e f f} = \frac{2 K_{1} + K_{2}}{3}$

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Space between plates is filled by two dielectrics K1 & K2, 23rd part is filled by K1 & 13rd is filled by K2, find effective dielectric constant. Area of the plates is A.

The correct option is A 2K1+K23 Both capacitors connected parallel. So Ceff=C1+C2 =ϵ02A3K1d+ϵ0A3K2d (ϵ0Ad)Keff=ϵ0Ad(2K13+K23) Keff=2K1+K23

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