(B) Given: Thickness of mica $d_1 = 1 \times 10^{-3} \,m$, thickness of fiber $d_2 = 0.5 \times 10^{-3} \,m$. Dielectric constants $K_1 = 8$ (mica) an

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(B) Given: Thickness of mica $d_1 = 1 \times 10^{-3} \,m$, thickness of fiber $d_2 = 0.5 \times 10^{-3} \,m$. Dielectric constants $K_1 = 8$ (mica) and $K_2 = 2.5$ (fiber). Breakdown electric field of fiber $E_2 = 6.4 \times 10^6 \,V/m$.Since the sheets are in series, the electric displacement field $D = \epsilon_0 K E$ is constant across the plates, so $K_1 E_1 = K_2 E_2$.Thus, $E_1 = E_2 (K_2 / K_1) = (6.4 \times 10^6) \times (2.5 / 8) = 2.0 \times 10^6 \,V/m$.The maximum voltage $V_{max}$ is given by $V_{max} = E_1 d_1 + E_2 d_2$.$V_{max} = (2.0 \times 10^6 \times 1 \times 10^{-3}) + (6.4 \times 10^6 \times 0.5 \times 10^{-3})$.$V_{max} = 2000 + 3200 = 5200 \,V$.

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

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The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

AP EAMCET 2023 All AP EAMCET

A
$3400$
B
$5200$
C
$2700$
D
$4800$

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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 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:

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The capacity of an air-filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric of constant $K$ as shown in the figure. The new capacity becomes $C_n$. The ratio of $C_n$ to $C_0$ is:

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In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

Medium

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$A$ parallel plate capacitor with plate separation $5 \text{ mm}$ is charged by a battery. On introducing a mica sheet of $2 \text{ mm}$ thickness and maintaining the connections of the plates with the terminals of the battery,it is found that it draws $25 \%$ more charge from the battery. The dielectric constant of mica is . . . . . . .

DifficultJEE MAIN 2026

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The capacity of an air-filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric of constant $K$ as shown in the figure. The new capacity becomes $C_n$. The ratio of $C_n$ to $C_0$ is:

In a capacitor of capacitance $20\,\mu F$,the distance between the plates is $2\,mm$. If a dielectric slab of width $1\,mm$ and dielectric constant $2$ is inserted between the plates,then the new capacitance is......$\mu F$.

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