Question 526: Electrostatics & Capacitance

Question

A parallel plate capacitor is charged by a battery and battery remains connected. The dielectric slab of constant '$$\mathrm{K}$$' is inserted between the plates and then taken out. Then electric field between the plates

Accepted Answer

Initial energy of the combined system
$$\mathrm{U}_1=\frac{1}{2} \mathrm{CV}_1^2+\frac{1}{2} \mathrm{CV}_2^2=\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2ight)$$
On joining the two condensers in parallel, common potential,
$$\mathrm{V}=\frac{\mathrm{V}_1+\mathrm{V}_2}{2}$$
$$	herefore \quad$$ Final energy of the combined system:
$$\mathrm{U}_2=\frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}ight)^2$$
$$	herefore \quad$$ Decrease in energy will be:
$$\begin{aligned}
\Delta \mathrm{U} & =\mathrm{U}_1-\mathrm{U}_2 \
& =\frac{\mathrm{C}}{2}\left(\mathrm{~V}_1^2+\mathrm{V}_2^2ight)-\frac{1}{2}(\mathrm{C}+\mathrm{C})\left(\frac{\mathrm{V}_1+\mathrm{V}_2}{2}ight)^2 \
& =\frac{1}{4} \mathrm{C}\left(\mathrm{V}_{\mathrm{f}}-\mathrm{V}_2ight)^2
\end{aligned}$$

Answer

remains the same

Answer

increases

Answer

decreases

Answer

becomes zero

Electrostatics & Capacitance

A parallel plate capacitor is charged by a battery and battery remains connected. The dielectric slab of constant '$\mathrm{K}$' is inserted between the plates and then taken out. Then electric field between the plates

Choose the correct answer: