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 $\text{ nC}$)

Three infinitely long linear charges of charge density $\lambda$,$\lambda$,and $-2\lambda$ are placed in space. $A$ point in space is specified by its perpendicular distances $r_1$,$r_2$,and $r_3$ respectively from the linear charges. For the points which are equipotential:

Two equal charges $q$ are placed at a distance of $2a$ and a third charge $-2q$ is placed at the midpoint. The potential energy of the system is:

Two charges $-q$ each are separated by a distance $2d$. $A$ third charge $+q$ is kept at the midpoint $O$. Find the potential energy of $+q$ as a function of a small distance $x$ from $O$ due to the $-q$ charges. Sketch the $P.E.$ versus $x$ graph and convince yourself that the charge at $O$ is in an unstable equilibrium.

In bringing an electron towards another electron,the electrostatic potential energy of the system

An electric dipole of dipole moment $6 \times 10^{-6} \ Cm$ is placed in a uniform electric field of magnitude $10^6 \ V/m$. Initially,the dipole moment is parallel to the electric field. The work that needs to be done on the dipole to make its dipole moment opposite to the field will be . . . . . . $J$.