Thank you for reporting, we will resolve it shortly
Q.
Two equal charges $q$ are placed at a distance of $2 a$ and a third charge $-2 q$ is placed at the midpoint. The potential energy of the system is
Electrostatic Potential and Capacitance
Solution:
$U=U_{1}+U_{2}+U_{3}$
$U =\frac{ K ( q )(-2 q )}{ a }+\frac{ K (-2 q )( q )}{ a }+\frac{ K ( q )( q )}{2 a }$
$U =\frac{-4 Kq ^{2}}{ a }+\frac{ Kq ^{2}}{2 a }$
$U =\frac{ Kq ^{2}}{ a }\left[-4+\frac{1}{2}\right]$
$U =\frac{ Kq ^{2}}{ a }\left(\frac{-7}{2}\right)$
$\Rightarrow U =\frac{-7 Kq ^{2}}{2 a }$
$U =\frac{-7 q ^{2}}{8 \pi \varepsilon_{ 0 }}a$