A capacitor of capacity $C_{0}$ is connected to a battery of emf $V_{0}$ . When steady is attained a dielectric slab of dielectric constant $K$ is slowly introduced in the capacitor. Find the force of attraction between the plates of the capacitor.

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Solution

When dielectric constant is introduced $C_{0}$ becomes ${K C}_{0}$
$q = K C_{0} V_{0}$
$F = \frac{Q^{2}}{2 A ε_{0} K} = \frac{K^{2} C_{0}^{2} V_{0}^{2}}{2 A ε_{0} K}$

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A capacitor of capacity C0 is connected to a battery of emf V0. When steady is attained a dielectric slab of dielectric constant K is slowly introduced in the capacitor. Find the force of attraction between the plates of the capacitor.

When dielectric constant is introduced C0 becomes KC0q=KC0V0F=Q22Aε0K=K2C20V202Aε0K

A capacitor of capacity $C_{0}$ is connected to a battery of emf $V_{0}$ . When steady is attained a dielectric slab of dielectric constant $K$ is slowly introduced in the capacitor. Find the force of attraction between the plates of the capacitor.

When dielectric constant is introduced $C_{0}$ becomes ${K C}_{0}$
$q = K C_{0} V_{0}$
$F = \frac{Q^{2}}{2 A ε_{0} K} = \frac{K^{2} C_{0}^{2} V_{0}^{2}}{2 A ε_{0} K}$