(D) $1$. When a dielectric slab of constant $K$ is present,the capacitance is $C_1 = \frac{K \varepsilon_0 A}{d}$.$2$. Since the capacitor remains con

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(D) $1$. When a dielectric slab of constant $K$ is present,the capacitance is $C_1 = \frac{K \varepsilon_0 A}{d}$.$2$. Since the capacitor remains connected to the battery,the potential difference across the plates remains constant,so $V_1 = V_2 = V$.$3$. When the dielectric slab is removed,the new capacitance becomes $C_2 = \frac{\varepsilon_0 A}{d}$.$4$. Since $K > 1$,it is clear that $C_1 > C_2$.$5$. Because the battery is still connected,the potential difference across the plates does not change,thus $V_1 = V_2$.

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

Medium

$A$ parallel plate capacitor of capacitance $C_1$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $V_1$ across its plates. When the dielectric slab is removed,keeping the capacitor connected to the battery,the new capacitance and potential difference are $C_2$ and $V_2$ respectively. Then:

KCET 2023 All KCET

A
$V_1 = V_2, C_1 < C_2$
B
$V_1 > V_2, C_1 > C_2$
C
$V_1 < V_2, C_1 > C_2$
D
$V_1 = V_2, C_1 > C_2$

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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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In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

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$A$ capacitor of capacity $C$ is connected to a cell of $V \, \text{volt}$. Now,a dielectric slab of dielectric constant $\epsilon_r$ is inserted into it while keeping the cell connected. Then:

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$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

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The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.

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In the figure,a capacitor is filled with dielectrics of constants $K_1$,$K_2$,and $K_3$. The resultant capacitance is

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The capacity of a parallel plate capacitor with no dielectric substance but with a separation of $0.4 \,cm$ is $2 \,\mu F$. The separation is reduced to half and it is filled with a dielectric substance of value $2.8$. The final capacity of the capacitor is.......$\mu F$.

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