(B) The initial capacitance of the capacitor is $C_{0} = \frac{\varepsilon_{0} A}{d}$.The initial charge on the capacitor is $q_{0} = C_{0} V = \frac{

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$\frac{k \varepsilon_{0} A}{d} V$

(B) The initial capacitance of the capacitor is $C_{0} = \frac{\varepsilon_{0} A}{d}$.The initial charge on the capacitor is $q_{0} = C_{0} V = \frac{\varepsilon_{0} A}{d} V$.When a dielectric slab of dielectric constant $k$ is inserted between the plates while the battery remains connected,the potential difference $V$ across the plates remains constant.The new capacitance becomes $C = k C_{0} = \frac{k \varepsilon_{0} A}{d}$.The final charge on the capacitor is $q = C V = \left( \frac{k \varepsilon_{0} A}{d} \right) V = k q_{0}$.

Class 12 Physics Electric Potential and Capacitance Effect of Dielectric Inside Capacitor

Medium

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

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A
$\frac{\varepsilon_{0} A}{d} V$
B
$\frac{k \varepsilon_{0} A}{d} V$
C
$\frac{\varepsilon_{0} A}{k d} V$
D
Zero

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ parallel plate capacitor has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

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$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

Similar Questions

The capacity of a parallel plate capacitor is $10\,\mu F$ without a dielectric. If a dielectric of constant $K = 2$ is used to fill half the distance between the plates,the new capacitance in $\mu F$ is:

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. The graph between electric field $(E)$ and distance $(x)$ from the left plate will be:

If a dielectric substance is introduced between the plates of a charged air-gap capacitor,the energy of the capacitor will:

$A$ parallel plate capacitor has a capacitance of $10 \ \mu F$ with air between its plates. Now,half of the space between the two plates is filled with a dielectric material of dielectric constant $K = 4$ as shown in the figure. Find the capacitance of the capacitor in $\mu F$.

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