Find the expression for the potential energy of a system of two point charges `q_(1), and q_(2)` glocated at `vecr_(1) and vecr_(2)` respectively in an external electric field `vecE.`
(b) Draw equipotential surfaces due to an isolated point charge (-q) and depict the electric field lines.
(c) Three point charges `+1muC,-1 muC and +2muC` are initially infinite distance apart. Calculate the work done in assembling these charges at the vertices of an equilateral triangle of side 10 cm.

Explanation

(a) To find the potential energy of a system of two point charges `vecr_(1) and vecr_(2)` situated at points and respectively in an external electric field `vecE`, first we calculate the work done in bringing the charge `q_(1)`. from infinity to `vecr_(1)`. Work done in this step is
`W_1=q_1V(vecr_(1))," where "V(vecr_1)` is the electrostatic potential at `vecr_(1)`.
Now we consider the work done in bringing `q_(2)" to "vecr_(2)`. In this step work W, is done against the electric field and work `W_2` is done against the field due to charge `q_1.` Obviously
`W_(2)=q_(2)V(vecr_(2))," where "V(vecr_(2))` is the electrostatic potential at `vecr_(2)`
and `W_(12)=(q_(1)q_(2))/(4pi epsi_(0).r_(12)) "where "r_(12)=|vecr_(1)-vecr_(2)|` = distance between `q_(1) and q_2`
Total work done in bringing `q_(2)" to "vecr_(2)=W_(2)+W_(12)=q_(2) V(vecr_(2))+(q_(1)q_(2))/(4pi in_(0).r_(12))`
Potential energy of the system U.
= The total work done in assembling the configuration
`U=W_(1)+W_(2)+W_(12)=q_(1) V(r_(1))+q_(2)V(r_(2))+(q_(1)q_(2))/(4pi epsi_(0).r_(12))`
(b) Equipotential surfaces due to an isolated point charge-q are drawn here and electric field lines have been depicted.
(c) Hence `q_(1)=+1 muC=+1 xx 10^(-6) C, q_(2)=-1muC=-1 xx 10^(-6) C, q_(3)=+2muC=+2 xx 10^(-6) C and r_(12)=r_(13)=r_(23)=10 cm =0.1m`
Work done is assembling the charges =Potential energy of charge alignment
`W=U=1/(4pi epsi_(0)) [(q_(1)q_(2))/(r_(12))+(q_(1)q_(3))/(r_(13))+(q_(2)q_(3))/(r_(23))]`

`=9 xx 10^(9) [(1 xx 10^(-6)) xx (-1 xx 10^(-6))/(0.1) +(1 xx 10^(-6)) xx (2 xx 10^(-6))/(0.1) +(-1 xx 10^(-6)) xx (2 xx 10^(-6))/(0.1)]`
=0.09J