(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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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $k(x)=K(1+\alpha x)$ where $x$ is the distance measured from one of the plates. If $(\alpha d) << 1$,the total capacitance of the system is best given by the expression:

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$A$ parallel plate capacitor has plate area $40 \text{ cm}^2$ and plate separation $2 \text{ mm}$. The space between the plates is filled with a dielectric medium of thickness $1 \text{ mm}$ and dielectric constant $K = 5$. The capacitance of the system is:

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Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

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$A$ parallel plate air capacitor is charged up to $100 \,V$. $A$ plate $2 \,mm$ thick is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6 \,mm$. The dielectric constant of the thick plate is

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$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

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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 has plate area $40 \text{ cm}^2$ and plate separation $2 \text{ mm}$. The space between the plates is filled with a dielectric medium of thickness $1 \text{ mm}$ and dielectric constant $K = 5$. The capacitance of the system is:

Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. $V$. If one of the capacitors is inserted in a liquid of dielectric constant $K$, then the potential difference across the other capacitor will become:

$A$ parallel plate air capacitor is charged up to $100 \,V$. $A$ plate $2 \,mm$ thick is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by $1.6 \,mm$. The dielectric constant of the thick plate is

$A$ capacitor has two circular plates,each with a radius of $8 \ cm$,separated by a distance of $1 \ mm$. Calculate the capacitance of this capacitor when a dielectric slab (dielectric constant $K = 6$) is placed between the plates.

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