Two charges $\mathit{q}\mathbf{=}\mathbf{+}\mathbf{\text{2.0μC}}$ are fixed a distance $\mathit{d}\mathbf{=}\mathbf{\text{2.0cm}}$apart (Fig. 24-69).

(a) With $\mathit{V}\mathbf{=}\mathbf{0}$at infinity, what is the electric potential at point C?

(b) You bring a third charge$\mathit{q}\mathbf{=}\mathbf{+}\mathbf{\text{2.0μC}}$ from infinity to C. How much work must you do?

(c) What is the potential energy U of the three-charge configuration when the third charge is in place?

The Coulomb’s constant, $k=9.0\times {10}^{\text{9}}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$

Th charge, $q=2.0\times {10}^{-6}\text{C}$

The distance, $d=2.0\times {10}^{-2}\text{m}$

After reading the question, the potential due to the chargeqat a distanceris,

$\mathbf{V}\mathbf{=}\frac{\mathrm{kq}}{r}$
Here, kis the electrostatic constant.

Below figure shows the arrangement of the two charges.

The distance between the two charges is d.

Formulae:

Write the equation for work done.

$W=q{V}_C$

The electric potential is define by using following formula.

$V_1=\frac{kq}{r_1}$

Where,W is work done,$r_1$is the distancebetween charge 1 and the point C ,and $V_C$is the potential energyto point C .

The net electric potential at point C is the sum of the electric potentials due to two charges.

$V_C=V_1+V_2$ ….. (1)

Here,$V_1$and$V_2$are the electric potentials due to the charges 1 and 2.

Use Pythagorean formula, the distance between charge 1 and the point C is,

$\begin{array}{c}{r}_1=\sqrt{{\left(\frac{d}{2}\right)}^2+{\left(\frac{d}{2}\right)}^2}\\ =\frac{d}{\sqrt{2}}\end{array}$

Hence, the potential due to the charge 1 at the point C is,

$\begin{array}{c}{V}_1=\frac{kq}{r_1}\\ =\frac{kq}{\frac{d}{\sqrt{2}}}\\ =\frac{\sqrt{2}kq}{d}\end{array}$

Similarly, the distance between the point C and the charge 2 is same as the distance between the charge 1 and point C.

Hence, the potential due to the charge 2 on the point C is,

$V_2=\frac{\sqrt{2}kq}{d}$

Convert the units for the charge from $\text{μC}$ to C .

$\begin{array}{c}q=\text{2.0μC}\\ =\text{2.0μC}\left(\frac{\text{1C}}{\text{1}\times {10}^6\text{μC}}\right)\\ =\text{2}.0\times {10}^{\text{-6}}\text{C}\end{array}$

Convert the units for the distance between the two charges from centimeter to meter.

$\begin{array}{c}d=\text{2.0cm}\\ =\text{2.0cm}\left(\frac{\text{1m}}{\text{100cm}}\right)\\ =2.0\times {10}^{-2}\text{m}\end{array}$

Substitute $\frac{\sqrt{2}kq}{d}$for $V_1$and $\frac{\sqrt{2}kq}{d}$for$V_2$ in the equation $\left(1\right)$, and solve by substituting known values,

$\begin{array}{c}{V}_C=\frac{\sqrt{2}kq}{d}+\frac{\sqrt{2}kq}{d}=2\left(\frac{\sqrt{2}kq}{d}\right)\\ =2\left(\frac{\sqrt{\text{2}}(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}})(2.0\times {10}^{-6}\text{C})}{2.0\times {10}^{-2}\text{m}}\right)\\ =2.54\times {10}^6\text{V}\end{array}$

Rounding off to two significant figures, the electric potential at the point C is $2.5\times {10}^6\text{V}$.

The work done for moving charge from infinity distance to the point C is,

$W=q{V}_C$

Here, q is the moved charge from infinity distance to point C.

Substitute $2.0\times {10}^{-6}\text{C}$ for q and ${\text{2.54×10}}^{\text{6}}\text{V}$ for $V_C$in the above equation.

$\begin{array}{c}W=(2.0\times {10}^{-6}\text{C})(2.54\times {10}^{\text{6}}\text{V})\\ =5.1\text{J}\end{array}$

Therefore, the required work done to move the charge from infinity to the point C is $\text{5.1J}$.

The net potential energy is sum of the potential energy between the two charges and the work done by moving charge from infinity distance to point C.

$U=U_{12}+W$ ….. (2)

Here,$U_{12}$is the potential energy between the two charges andis the work done for bringing charge from infinity distance to point C.

The potential energy between the two charges is,

$U_{12}=\frac{kqq}{d}$

Substitute$\frac{kqq}{d}$ for$U_{12}$in the equation$\left(2\right)$and solve for U .

$U=\frac{kqq}{d}+W$

Substitute$9.0\times {10}^{\text{9}}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}$for K , $2.0\times {10}^{-6}\text{C}$for q, $\text{5.1J}$forW , and$2.0\times {10}^{\text{-2}}\text{m}$for d in the above equation.

$\begin{array}{c}U=\frac{(9.0\times {10}^9\text{N}\cdot {\text{m}}^{\text{2}}/{\text{C}}^{\text{2}}){(2.0\times {10}^{-6}\text{C})}^{\text{2}}}{2.0\times {10}^{-2}\text{m}}+5.1\text{J}\\ =1.8\text{J}++5.1\text{J}\\ =\text{6.9J}\end{array}$

Hence, potential energy of the three charge configuration is $\text{6.9J}$.