The figure shows a part of a circuit. The voltage across A and B is 4V. Then the voltages across 2 $\mu$F and 1.5 $\mu$F are respectively
Two equipotential surfaces A and B are separated by a distance x. The work done in moving a charge -q from A to B is (Assume $\varepsilon_0$ = permittivity of free space)
In the given circuit, the potential difference between B and D is zero. Then the value of the current I is
Two mercury drops of radii $ r $ and $ 2r $ merge to form a bigger drop. The surface energy released in the process is nearly (Surface tension of mercury is } $S$ and take $9^{2/3} = 4.326$)