The work done to assemble the three charges in a configuration as shown in the figure below is

Question

The work done to assemble the three charges in a configuration as shown in the figure below is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/8662838550ff2b0c1323c61439d55b2c-.png" /> (A) (-3 q2/4 π ε0 a) (B) (-2 q2/4 π ε0 a) (C) (-q2/4 π ε0 a) (D) 0

Tardigrade Admin · Accepted Answer

Work done, W=U12+U23+U13 =(k q1 q2/r12)+(k q2 k q3/r23)+(k q1 q3/r13) =(1/4 π ε0 a)[-2 q2+2 q2-q2] =-(q2/4 π ε0 a)

Q. The work done to assemble the three charges in a configuration as shown in the figure below is

$\frac{- 3 q ^{2}}{4 π ε _{0} a}$

$\frac{- 2 q ^{2}}{4 π ε _{0} a}$

$\frac{- q ^{2}}{4 π ε _{0} a}$

$0$

Work done,
$W = U_{12} + U_{23} + U_{13}$
$= \frac{k q _{1} q _{2}}{r _{12}} + \frac{k q _{2} k q _{3}}{r _{23}} + \frac{k q _{1} q _{3}}{r _{13}}$
$= \frac{1}{4 π ε _{0} a} [- 2 q^{2} + 2 q^{2} - q^{2}]$
$= - \frac{q ^{2}}{4 π ε _{0} a}$

Q. The work done to assemble the three charges in a configuration as shown in the figure below is

4πε0​a−3q2​

4πε0​a−2q2​

4πε0​a−q2​

0

Work done, W=U12​+U23​+U13​ =r12​kq1​q2​​+r23​kq2​kq3​​+r13​kq1​q3​​ =4πε0​a1​[−2q2+2q2−q2] =−4πε0​aq2​

$\frac{- 3 q ^{2}}{4 π ε _{0} a}$

$\frac{- 2 q ^{2}}{4 π ε _{0} a}$

$\frac{- q ^{2}}{4 π ε _{0} a}$

$0$

Work done,
$W = U_{12} + U_{23} + U_{13}$
$= \frac{k q _{1} q _{2}}{r _{12}} + \frac{k q _{2} k q _{3}}{r _{23}} + \frac{k q _{1} q _{3}}{r _{13}}$
$= \frac{1}{4 π ε _{0} a} [- 2 q^{2} + 2 q^{2} - q^{2}]$
$= - \frac{q ^{2}}{4 π ε _{0} a}$