Question

A point charge $q_1=-5.8\mu \text{C}$ is held stationary at the origin. A second charge $q_2=+4.3\mu \text{C}$ moves from the point $(0.26\text{m},0,0)$ to $(0.38\text{m},0,0)$. Find the workdone by the electric force on $q_2$.

• A. $-0.272\text{J}$
• B. $+0.272\text{J}$
• C. $-0.381\text{J}$
• D. $+0.381\text{J}$

Answer 1

The correct option is A $-0.272\text{J}$

Given,

$q_1=-5.8\mu \text{C},q_2=+4.3\mu \text{C}$

$r_i=0.26\text{m},r_f=0.38\text{m}$

Workdone by the electrostatic force $W=U_i-U_f$

$\because \Delta U=\frac{q_1{q}_2}{4\pi {\epsilon }_0}(\frac{1}{r_f}-\frac{1}{r_i})$

We can write that,

$W=\frac{q_1{q}_2}{4\pi {\epsilon }_0}(\frac{1}{r_i}-\frac{1}{r_f})$

Substituting the given data we get,

$W=\frac{(-5.8\times {10}^{-6})(4.3\times {10}^{-6})(9\times {10}^9)(0.38-0.26)}{0.38\times 0.26}$

$\Rightarrow W=-0.272\text{J}$

Hence, option (a) is the correct answer.