(B) When a dielectric of constant $K$ is introduced between the plates of a parallel plate capacitor,its capacity increases by a factor of $K$. Theref

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(B) When a dielectric of constant $K$ is introduced between the plates of a parallel plate capacitor,its capacity increases by a factor of $K$. Therefore,the new capacity $C' = K \times C_0 = 5C_0$.Assuming the charge $q$ on the plates remains constant (isolated capacitor),the energy stored is given by $W = \frac{q^2}{2C}$.The initial energy is $W_0 = \frac{q^2}{2C_0}$.The new energy $W'$ is given by $W' = \frac{q^2}{2C'} = \frac{q^2}{2(5C_0)} = \frac{1}{5} \left( \frac{q^2}{2C_0} \right) = \frac{W_0}{5}$.Thus,the new capacity is $5C_0$ and the new energy is $\frac{W_0}{5}$.

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

Medium

The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_0$ and $W_0$. If the air is replaced by glass (dielectric constant $K = 5$) between the plates,the capacity of the plates and the energy stored in it will respectively be

A
$5C_0, 5W_0$
B
$5C_0, \frac{W_0}{5}$
C
$\frac{C_0}{5}, 5W_0$
D
$\frac{C_0}{5}, \frac{W_0}{5}$

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

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$A$ parallel plate capacitor has plates with area $A$ and separation $d$. $A$ battery charges the plates to a potential difference $V_0$. The battery is then disconnected and a dielectric slab of thickness $d$ is introduced. The ratio of energy stored in the capacitor before and after the slab is introduced is

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$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

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Consider the arrangement shown in the figure. The total energy stored is $U_1$ when the key $K$ is closed. Now,the key $K$ is opened,and two dielectric slabs of relative permittivity $\epsilon_r$ are introduced between the plates of the two capacitors. The slabs fit tightly between the plates. The total energy stored is now $U_2$. Then,the ratio $U_1/U_2$ is:

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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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The capacity and the energy stored in a parallel plate condenser with air between its plates are respectively $C_0$ and $W_0$. If the air is replaced by glass (dielectric constant $K = 5$) between the plates,the capacity of the plates and the energy stored in it will respectively be

Similar Questions

$A$ parallel plate capacitor with oil between the plates (dielectric constant of oil $K = 2$) has a capacitance $C$. If the oil is removed,then the capacitance of the capacitor becomes:

$A$ parallel plate capacitor with air between the plates has a capacitance of $8 \;pF$ $(1 \;pF = 10^{-12} \;F)$. What will be the capacitance (in $pF$) if the distance between the plates is reduced by half,and the space between them is filled with a substance of dielectric constant $6$?

$A$ 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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