Question 547: Electrostatics & Capacitance

Question

The potential energy of charged parallel plate capacitor is $$v_0$$. If a slab of dielectric constant $$\mathrm{K}$$ is inserted between the plates, then the new potential energy will be

Accepted Answer

We know, $$v_0=\frac{Q^2}{2 C}$$
On inserting the slab of dielectric constant $$\mathrm{k}$$, the new capacitance $$\mathrm{C}^{\prime}=\mathrm{KC}$$
$$	herefore \quad$$ New potential energy $$v_0^{\prime}=\frac{Q^2}{2 C^{\prime}}$$
$$v_0^1=\frac{\mathrm{Q}^2}{2 \mathrm{KC}}=\frac{v_0}{\mathrm{~K}}$$

Answer

$$\frac{v_0}{\mathrm{~K}}$$

Answer

$$v_0 \mathrm{~K}^2$$

Answer

$$\frac{v_0}{\mathrm{~K}^2}$$

Answer

$$\mathrm{v}_0^2$$

Electrostatics & Capacitance

The potential energy of charged parallel plate capacitor is $v_0. If a slab of dielectric constant \mathrm{K}$ is inserted between the plates, then the new potential energy will be

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