If ${10}^9$ electrons move out of a body to another body every second, then the time required to get a total charge of $1\mathrm{ }\mathrm{C}$ on the other body is

Two charges of equal magnitudes and at a distance $r$ exert a force $F$ on each other. If the charges are halved and distance between them is doubled, then the new force acting on each charge is

Four point charges $q_A=2 \mu C, q_B=-5 \mu C, q_C=2 \mu C, q_D=-5 \mu C$ are located at the corners of a square $ABCD$ of side $10 \mathrm{cm}.$ The force acting on a charge of $1 \mu \mathrm{C}$ placed at centre of the square is

Two equal point charges of $1 \mu \mathrm{C}$ each are located at points $(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \mathrm{m}$ and $(2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-\hat{\mathrm{k}}) \mathrm{m}$. What is the magnitude of electrostatic force between them?

As shown in the figure, two particles, each of mass $\text{'}m\text{'}$ tied at the ends of a light string of length $2a$ are kept on a frictionless horizontal surface. When the mid point $\left(P\right)$ of the string is pulled vertically upwards with a small but constant force $F,$ the particles move towards each other on the surface. Magnitude of acceleration of each particle, when the separation between them becomes $2x$ is