91Ó°ÊÓ

1. Length of the rectangle$\mathrm{L}=15\mathrm{cm}\mathrm{or}0.15\mathrm{m}$
2. Width of the rectangle$\mathrm{w}=5\mathrm{cm}\mathrm{or}0.05\mathrm{m}$
3. Charge at the top corner${\mathrm{q}}_{\mathrm{t}}=-5.0×{10}^{-6}\mathrm{C}$
4. Charge at the bottom corner${\mathrm{q}}_2=2.0×{10}^{-6}\mathrm{C}$
5. Charge at the bottom corner${\mathrm{q}}_3=3.0×{10}^{-6}\mathrm{C}$
6. With V=0 at infinity.

Using the formula of the electric potential of a system, we can the desired values of the electric potentials at corners A and B. Again, this value will determine the potential energy at the corners and the difference value gives the work required to move the third particle. Again, as the work is conservative, it is independent of all the paths taken by the charges.

Formulae:

The electric potential at a point, $\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{r}}$ (i)

The potential energy of the system in terms of potential, U=qV (ii)

The total electric potential at the corner is due to the charges present along the length and the width. So using the given data and equation (i), we get the potential as:

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{A}}& =& \mathrm{k}\left(\frac{{\mathrm{q}}_1}{\mathrm{L}}+\frac{{\mathrm{q}}_2}{\mathrm{W}}\right)\\ & =& \left(8.99×{10}^9{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)\left(\frac{-5.0×{10}^{-6}\mathrm{C}}{0.15\mathrm{m}}+\frac{20×{10}^{-6}\mathrm{C}}{0.05\mathrm{m}}\right)\\ & =& 59.93×{10}^3\mathrm{V}\\ & ≈& 60.0×{10}^3\mathrm{V}\end{array}$

Hence, the value of the electric potential is $60.0×{10}^3\mathrm{V}$.

Similarly, the electric potential at corner B is given using the given data in equation (i) as follows:

$\begin{array}{rcl}{\mathrm{V}}_{\mathrm{B}}& =& \mathrm{k}\left(\frac{{\mathrm{q}}_1}{\mathrm{W}}+\frac{{\mathrm{q}}_2}{\mathrm{L}}\right)\\ & =& \left(8.99×{10}^8{\mathrm{Nm}}^2/{\mathrm{C}}^2\right)\left(\frac{-5.0×{10}^{-6}\mathrm{C}}{0.05\mathrm{m}}+\frac{2.0×{10}^{-6}\mathrm{C}}{0.15\mathrm{m}}\right)\\ & =& -7.8×{10}^5\mathrm{V}\end{array}$

Hence, the value of the potential is data-custom-editor="chemistry" $-7.8×{10}^5\mathrm{V}$.

Work required moving a charge q from B to A be equal to difference in potential energy between point A and point B. So, using the above values, the work done is given using equation (ii) as follows:

$\begin{array}{rcl}\mathrm{W}& =& {\mathrm{U}}_{\mathrm{A}}-{\mathrm{U}}_{\mathrm{B}}\\ & =& {\mathrm{q}}_3\left({\mathrm{V}}_{\mathrm{A}}-{\mathrm{V}}_{\mathrm{B}}\right)\\ & =& \left(3.0×{10}^{-6}\mathrm{C}\right)\left(60.0×{10}^3\mathrm{V}-\left(-7.8×{10}^5\mathrm{V}\right)\right)\\ & =& 2.52\mathrm{J}\end{array}$

Hence, the value of the work is 2.52 J.

Since the work done by the external agent is positive. So this work increases the electric potential energy of the three-charge system.

The work done depends only on initial and final positions only. So the work done is independent of path. Hence the work done is same on all paths between two points.

The work done depends only on initial and final positions only. So the work done is independent of path. Hence the work done is same on all paths between two points.