Two particles of charges $4 \text{ nC}$ and $Q$ are kept in air with a separation of $10 \text{ cm}$ between them. If the electrostatic potential energy of the system is $1.8 \mu \text{ J}$, then $Q$ is: (in $\text{ nC}$)

Two point charges $q_1 = 6 \mu\text{C}$ and $q_2 = 4 \mu\text{C}$ are kept at points $A$ and $B$ in air where distance $AB = 10 \text{cm}$. What is the increase in potential energy of the system when $q_2$ is moved towards $q_1$ by $2 \text{cm}$? $(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \text{SI units})$

Three point charges of $3 \mu C, 4 \mu C$,and $5 \mu C$ are arranged at the three corners of a right-angled triangle $ABC$ as shown in the figure. The work done in moving the charges at $A$ and $C$,so that the three charges are located at the three corners of an equilateral triangle of side $3 \text{ cm}$ is (in $J$)

In the figure shown,the electric potential energy of the system is: ($q$ is at the centre of the conducting neutral spherical shell of inner radius $a$ and outer radius $b$)

The work done required to assemble the four charges at the corners of a square of side $a$, as shown in the figure, is:

Two positive point charges of $10 \ \mu C$ and $12 \ \mu C$ are placed $10 \ cm$ apart in air. The work done to bring them $6 \ cm$ closer is (in $J$)