Question 261:
easyEight equally charged tiny drops are combined to form a big drop. If the potential on each drop is $10 \text{ V}$ then potential of big drop will be: (1999)
By conservation of volume, $8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \implies R = 2r$. Charge is conserved, so $Q = 8q$. The new potential $V' = \frac{kQ}{R} = \frac{k(8q)}{2r} = 4 \left(\frac{kq}{r}\right) = 4 \times 10 \text{ V} = 40 \text{ V}$.
Question 262:
easyThere is an electric field $E$ in x-direction. If the work done in moving a charge of $0.2 \text{ C}$ through a distance of $2 \text{ m}$ along a line making an angle $60^{\circ}$ with x-axis is $4 \text{ J}$, then what is the value of $E$? (1995)
Work done is given by $W = qEd \cos\theta$. Substituting the values: $4 = 0.2 \times E \times 2 \times \cos(60^{\circ})$. This simplifies to $4 = 0.4 \times E \times 0.5$, which gives $E = \frac{4}{0.2} = 20 \text{ N/C}$.
Question 263:
easyCharge $q_2$ is at the centre of a circular path with radius $r$. Work done in carrying charge $q_1$, once around this equipotential path, would be (1994)
A circular path around a point charge is an equipotential surface because the distance $r$ from the central charge $q_2$ is constant. Since the potential difference between any two points on this path is zero, the work done $W = q\Delta V$ is zero.