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Byju's Answer

Standard XII

Physics

Partially Filled Dielectrics

The capacity ...

Question

The capacity of a parallel plate condenser is $10 μ F$ without the dielectric. Material with a dielectric constant of $2$ is used to fill half-thickness between the plates. The new capacitance is $- ------- - μ F$ :

10

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20

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15

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13.33

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Solution

The correct option is B $13.33$
Initial capacitance $= \frac{ϵ_{0} A}{d} = 10 μ F$
Now effective capacitance is series combination of $\frac{ϵ_{0} A}{\frac{d}{2}}$ and $\frac{k ϵ_{0} A}{\frac{d}{2}}$
$∴ C_{e f f} =$ series of $20 μ F$ and $40 μ F$
$\frac{1}{C_{e f f}} = \frac{1}{20} + \frac{1}{40}$
$C_{e f f} = 13.33 μ F$

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The capacity of a parallel plate condenser is 10μF without the dielectric. Material with a dielectric constant of 2 is used to fill half-thickness between the plates. The new capacitance is –––––––––μF :

The correct option is B 13.33Initial capacitance =ϵ0Ad=10μFNow effective capacitance is series combination of ϵ0Ad2 and kϵ0Ad2∴Ceff= series of 20μF and 40μF1Ceff=120+140Ceff=13.33μF

The capacity of a parallel plate condenser is $10 μ F$ without the dielectric. Material with a dielectric constant of $2$ is used to fill half-thickness between the plates. The new capacitance is $- ------- - μ F$ :