Three charges $-q, Q$ and $-q$ are placed respectively at equal distances on a straight line. If the potential energy of the system of three charges is zero,then what is the ratio of $Q: q$ ?

Three particles,each having a charge of $10 \mu C$,are placed at the corners of an equilateral triangle of side $10 \ cm$. The electrostatic potential energy of the system is (Given $\frac{1}{4 \pi \varepsilon_{0}} = 9 \times 10^{9} \ N \cdot m^{2}/C^{2}$):

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 $J$)

$A$ moving positive charge approaches a negative charge. What will happen to the 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 . . . . . . $J$.

Three charges $Q, (+q)$ and $(+q)$ are placed at the vertices of an equilateral triangle of side $l$ as shown in the figure. If the net electrostatic potential energy of the system is zero,then $Q$ is equal to