(D) The potential difference $V$ across a capacitor is given by $V = E \cdot d$. When a dielectric slab of thickness $t$ is inserted,the new potential

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(D) The potential difference $V$ across a capacitor is given by $V = E \cdot d$. When a dielectric slab of thickness $t$ is inserted,the new potential difference $V'$ is $V' = E(d - t + t/K)$.To maintain the same voltage $V = V'$,we have $d = d - t + t/K$,which simplifies to $t(1 - 1/K) = d_{new} - d_{old}$.Here,the increase in distance is $d' = t(1 - 1/K)$.Rearranging for $K$: $d'/t = 1 - 1/K$,so $1/K = 1 - d'/t = (t - d')/t$.Thus,$K = t / (t - d')$.Given $t = 4 \times 10^{-5} \ m$ and $d' = 3.5 \times 10^{-5} \ m$:$K = (4 \times 10^{-5}) / (4 \times 10^{-5} - 3.5 \times 10^{-5})$$K = 4 / 0.5 = 8$.

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

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$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?

A
$10$
B
$12$
C
$6$
D
$8$

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Electrostatic Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor has a plate separation of $d$. $A$ metal plate of thickness $d/2$ and the same area as the capacitor plates is inserted between them without touching the plates. The capacitance of the resulting capacitor becomes:

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When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

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

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Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

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$A$ capacitor is charged. When a dielectric slab of thickness $t = 4 \times 10^{-5} \ m$ is inserted between the plates,the distance between the plates has to be increased by $d' = 3.5 \times 10^{-5} \ m$ to maintain the same voltage. What is the dielectric constant $K$ of the dielectric?

Similar Questions

$A$ parallel plate capacitor has a plate separation of $d$. $A$ metal plate of thickness $d/2$ and the same area as the capacitor plates is inserted between them without touching the plates. The capacitance of the resulting capacitor becomes:

When a dielectric substance is placed between the two plates of a capacitor,what happens to its capacitance,potential difference,and potential energy,respectively?

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.......$\mu F$.

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

Half of the space between the plates of a parallel-plate capacitor is filled with a dielectric material of dielectric constant $K$. The remaining half contains air. The capacitor is now given a charge $Q$. Then:

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