(B) The electric field between the plates of a capacitor in a vacuum or air is given by $E = \frac{\sigma}{\epsilon_0}$.When the capacitor is immersed

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Decreased proportional to $\frac{1}{2}$

(B) The electric field between the plates of a capacitor in a vacuum or air is given by $E = \frac{\sigma}{\epsilon_0}$.When the capacitor is immersed in a dielectric medium with dielectric constant $K$,the electric field is reduced by a factor of $K$.The new electric field $E_{medium}$ is given by $E_{medium} = \frac{E}{K}$.Given $K = 2$,the new electric field is $E_{medium} = \frac{E}{2}$.Therefore,the electric field is decreased proportional to $\frac{1}{2}$.

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

Easy

$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:

A
Increased proportional to $2$
B
Decreased proportional to $\frac{1}{2}$
C
Increased proportional to $\sqrt{2}$
D
Decreased proportional to $\frac{1}{\sqrt{2}}$

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Class 12 Questions

Class 12

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Chapter

Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

$A$ parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$,the dielectric constant of the medium is

Medium

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The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.

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$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $K = 5$ as shown in the figure,the percentage increase in the capacitance will be......$\%$

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The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used,is $C$. If the dielectric is replaced by another with dielectric constant $20$,the capacity will become

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$A$ parallel plate capacitor has two layers of dielectrics as shown in the figure. This capacitor is connected across a battery. The ratio of the potential difference across the dielectric layers $(V_1/V_2)$ is:

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$A$ parallel plate capacitor is immersed in an oil of dielectric constant $K = 2$. The electric field between the plates is:

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$A$ parallel plate capacitor has a capacity $C$. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes $2C$,the dielectric constant of the medium is

The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.

$A$ parallel plate air capacitor has a capacitance $C$. When it is half filled with a dielectric of dielectric constant $K = 5$ as shown in the figure,the percentage increase in the capacitance will be......$\%$

The capacity of a condenser in which a dielectric of dielectric constant $5$ has been used,is $C$. If the dielectric is replaced by another with dielectric constant $20$,the capacity will become

$A$ parallel plate capacitor has two layers of dielectrics as shown in the figure. This capacitor is connected across a battery. The ratio of the potential difference across the dielectric layers $(V_1/V_2)$ is:

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