(B) The capacitance of a parallel plate capacitor with air between the plates is given by $C = \frac{\epsilon_0 A}{d}$.When a conducting slab (copper)

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(B) The capacitance of a parallel plate capacitor with air between the plates is given by $C = \frac{\epsilon_0 A}{d}$.When a conducting slab (copper) of thickness $t$ is inserted between the plates,the new capacitance $C'$ is given by $C' = \frac{\epsilon_0 A}{d - t}$.Given that the thickness of the copper slab is $t = d/2$,we substitute this into the formula:$C' = \frac{\epsilon_0 A}{d - d/2} = \frac{\epsilon_0 A}{d/2} = 2 \left( \frac{\epsilon_0 A}{d} \right)$.Since $C = \frac{\epsilon_0 A}{d}$,we have $C' = 2C$.Therefore,the ratio $C'/C = 2$.

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

Medium

$A$ copper slab of thickness $d/2$ is inserted between the plates of a parallel plate capacitor,where $d$ is the distance between its two plates. If the capacitance of the capacitor without the copper slab is $C$ and with the copper slab is $C'$,find the ratio $C'/C$.

A
$\sqrt{2}$
B
$2$
C
$1$
D
$1/\sqrt{2}$

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Class 12

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Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

Medium

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The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

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$A$ parallel plate capacitor is made of two square plates of side $a$,separated by a distance $d$ $(d \ll a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$,as shown in the figure. The capacitance of this capacitor is

DifficultJEE MAIN 2019

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When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

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In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

MediumMHT CET 2019

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$A$ copper slab of thickness $d/2$ is inserted between the plates of a parallel plate capacitor,where $d$ is the distance between its two plates. If the capacitance of the capacitor without the copper slab is $C$ and with the copper slab is $C'$,find the ratio $C'/C$.

Similar Questions

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

The plates of a parallel plate capacitor with no dielectric are connected to a voltage source. Now,a dielectric of dielectric constant $K$ is inserted to fill the whole space between the plates,with the voltage source remaining connected to the capacitor.

$A$ parallel plate capacitor is made of two square plates of side $a$,separated by a distance $d$ $(d \ll a)$. The lower triangular portion is filled with a dielectric of dielectric constant $K$,as shown in the figure. The capacitance of this capacitor is

When a capacitor is filled with a dielectric constant $K = 3$,the charge is $Q_0$,voltage is $V_0$,and electric field is $E_0$. If the capacitor is now filled with a dielectric constant $K = 9$,what will be the new charge,voltage,and electric field respectively?

In a parallel plate air capacitor, the distance between plates is reduced to one-fourth and the space between them is filled with a dielectric medium of constant $2$. If the initial capacity of the capacitor is $4 \mu F$, then its new capacity is: (in $\mu F$)

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