Question

Q. A unit positive charge has to brought from infinity to a mid-point between two charges 20μC and 10μC separated by a distance of 50 m. How much work will be required ?
(a) 10.8 X 10⁴ J (b) 10.8 X 10³ J (c) 1.08 X 10⁶ J (d) 0.54 X 10⁵ J

Answer 1

Dear Student,



The work done by the two charges 20$\mu$C and 10​$\mu$C to bring the unit positive charge in mid point of the line joining them
will be stored as the electrical potential energy of the system.

$W=U=\frac{1}{4\pi {\epsilon }_0}\left(\frac{q1q2}{\left|r_2-r_1\right|}+\frac{q1q3}{\left|r_3-r_1\right|}\right)\left(1\right)$

q1 -unit positive test charge, q2 -20​$\mu$C,q3-10​$\mu$C and $\left|r_2-r_1\right|=\left|r_3-r_1\right|=25m$

$k=\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}Vm/C$ . Substituting the values in equation (1), we get


$U=9\times {10}^9\times \frac{6}{5}\times {10}^{-6}$
i.e., U = 10.8 $\times$10³ J .


Hence option B is the solution.




Regards