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
Two identical capacitors have the same capacitance '$$C$$'. One of them is charged to a potential $$V_1$$ and the other to $$V_2$$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is
If a capacitor of capacity $$900 ~\mu \mathrm{F}$$ is charged to $$100 \mathrm{~V}$$ and its total energy is transferred to a capacitor of capacity $$100 ~\mu \mathrm{F}$$, then its potential will be
Two dielectric slabs having dielectric constant '$$\mathrm{K}_1$$' and '$$\mathrm{K}_2$$' of thickness $$\frac{\mathrm{d}}{4}$$ and $$\frac{3 \mathrm{~d}}{4}$$ are inserted between the plates as shown in figure. The net capacitance between $$A$$ and $$B$$ is $$\left[\varepsilon_0\right.$$ is permittivity of free space]
