A parallel plate capacitor of capacitance $$C_1$$ with a dielectric slab in between its plates is connected to a battery. It has a potential difference $$V_1$$ across its plates. When the dielectric slab is removed, keeping the capacitor connected to the battery, the new capacitance and potential difference are $$C_2$$ and $$V_2$$ respectively, Then
A cubical Gaussian surface has side of length $$a=10 \mathrm{~cm}$$. Electric field lines are parallel to $$X$$-axis as shown in figure. The magnitudes of electric fields through surfaces $$A B C D$$ and $$E F G H$$ are $$6 ~\mathrm{kNC}^{-1}$$ and $$9 \mathrm{~kNC}^{-1}$$ respectively. Then, the total charge enclosed by the cube is
[Take, $$\varepsilon_0=9 \times 10^{-12} \mathrm{~Fm}^{-1}$$ ]
Electric field at a distance $$r$$ from an infinitely long uniformly charged straight conductor, having linear charge density $$\lambda$$ is $$E_1$$. Another uniformly charged conductor having same linear charge density $$\lambda$$ is bent into a semicircle of radius $$r$$. The electric field at its centre is $$E_2$$. Then
A wire of resistance $$R$$ is connected across a cell of emf $$(\varepsilon)$$ and internal resistance $$(r)$$. The current through the circuit is $$I$$. In time $$t$$, the work done by the battery to establish the current $$I$$ is