(A) The capacitance of a parallel plate capacitor is given by $C = \frac{K A \varepsilon_0}{d}$.Initially,for air,$K_1 = 1$ and $d_1 = d$,so $C_1 = \f

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(A) The capacitance of a parallel plate capacitor is given by $C = \frac{K A \varepsilon_0}{d}$.Initially,for air,$K_1 = 1$ and $d_1 = d$,so $C_1 = \frac{A \varepsilon_0}{d} = 12 \mu F$.When the distance is doubled,$d_2 = 2d$,and the dielectric constant is $K_2 = 4$.The new capacitance $C_2$ is given by $C_2 = \frac{K_2 A \varepsilon_0}{d_2}$.Substituting the values,we get $C_2 = \frac{4 A \varepsilon_0}{2d} = 2 \left( \frac{A \varepsilon_0}{d} \right) = 2 \times C_1$.Therefore,$C_2 = 2 \times 12 \mu F = 24 \mu F$.

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

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$A$ parallel plate capacitor with air between the plates has a capacitance of $12 \mu F$. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $4$,what will be the new capacitance of the capacitor (in $\mu F$)?

AP EAMCET 2022 All AP EAMCET

A
$24$
B
$72$
C
$6$
D
$12$

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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 with air between the plates has a capacitance of $12 \mu F$. If the distance between the plates is doubled and the space between the plates is filled with a substance of dielectric constant $4$,what will be the new capacitance of the capacitor (in $\mu F$)?

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