Question

Two point charges $q_1=-1\mu \text{C}$ and $q_2=+1\mu \text{C}$ are located on the $x-$ axis, at $x=0\text{m}$ and at $x=1\text{m}$ respectively. The work that must be done by an external force to bring a third point charge $q_3=+1\mu \text{C}$ from infinity to $x=2\text{m}$ is

• A. $1.5\times {10}^{-3}\text{J}$
• B. $4.5\times {10}^{-3}\text{J}$
• C. $2.5\times {10}^{-3}\text{J}$
• D. $3\times {10}^{-3}\text{J}$

Answer 1

The correct option is B $4.5\times {10}^{-3}\text{J}$

The work that must be done on $q_3$ by an external force is equal to the difference of potential energy $U$ when the charge is at $x=2\text{m}$ and the potential energy when it is at infinity.

$\therefore W=U_f-U_i$

$W=\frac{1}{4\pi {ϵ}_0}[\frac{q_3{q}_2}{(r_{32}{)}_f}+\frac{q_3{q}_1}{(r_{31}{)}_f}+\frac{q_2{q}_1}{(r_{21}{)}_f}]-\frac{1}{4\pi {ϵ}_0}[\frac{q_3{q}_2}{(r_{32}{)}_i}+\frac{q_3{q}_1}{(r_{31}{)}_i}+\frac{q_2{q}_1}{(r_{21}{)}_i}]$

Here, $(r_{21}{)}_i=(r_{21}{)}_f$

Also, $(r_{32}{)}_i=(r_{31}{)}_i=\infty$

$\therefore$ $W=\frac{1}{4\pi {ϵ}_0}[\frac{q_3{q}_2}{(r_{32}{)}_f}+\frac{q_3{q}_1}{(r_{31}{)}_f}]$

Substituting the values, we have

$W=(9.0\times {10}^9)\left({10}^{-12}\right)[\frac{\left(1\right)\left(1\right)}{\left(1.0\right)}+\frac{\left(1\right)(-1)}{\left(2.0\right)}]$

$W=4.5\times {10}^{-3}\text{J}$

Hence, option (b) is the correct answer.