91Ó°ÊÓ

Net force is calculated from the Coulomb’s law,

$F=k\frac{\left|q{q}^{\text{'}}\right|}{r^2}$

Here, the net force on q0is due to the force exerted by each charge at the opposite corners of the square. Refer to the image below.

Thus,

$$\begin{gathered} \mathrm{F}=\mathrm{k}\frac{\left(5.00\mathrm{μ°ä}\right)\left(3.00\mathrm{μ°ä}\right)}{{\left(0.0800\mathrm{m}\right)}^2} \\ =21.07\mathrm{N} \end{gathered}$$

Vector addition of the two forces gives the net force,

$$\begin{gathered} F=\left(21.07\mathrm{N}\right)\sqrt{2} \\ =29.8\mathrm{N} \end{gathered}$$

The direction of the force 29.8 N is from A to B as they both attract the test charge

When the point B is considered, the force exerted by both the charges are equal and opposite in direction and thus the vector addition of the forces on the test charge turns out to be zero.

Given data in the question is,

q0 = –3.00 μ°ä, q = 5.00 μ°ä, $r_a$= 0.0800 m, and $r_b$= 0.0400m.

The potential expression is,

$V=k\frac{q}{r}$

Use this in the expression of work done for both the charge together,

$$\begin{gathered} W_{FE}=2q\left(V_A-V_B\right) \\ =2kq{q}_0\left(\frac{1}{r_A}-\frac{1}{r_B}\right) \end{gathered}$$

Substitute all the values in the above equation and solve to get the value of the work done as

$\mathrm{W}=+1.40\mathrm{J}.$

Thus, the work +1.40 J is positive because the force drives the charge from point A to B in the direction of the field. As the charge q0 is negative, it moves from lower to higher potential.