In Fig. 24-70, point P is at the center of the rectangle. With $\mathbf{V}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$at infinity, ${\mathbf{q}}_{\mathbf{1}}\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{00}\mathbf{fC}$ , ${\mathit{q}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{00}\mathit{f}\mathit{C}$ , ${\mathbf{q}}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{fC}$, and$\mathbf{d}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{54}\mathbf{cm}$, what is the net electric potential at P due to the six charged particles?

Convert the units of charge from $\text{fC}$to $\text{C}$.

The charge,

$$\begin{gathered} q_1=\text{5 fC} \\ =\left(\text{5 fC}\right)\left(\frac{{10}^{-15}\text{C}}{\text{1.0 fC}}\right) \\ =5\times {10}^{-15}\text{C} \end{gathered}$$

The charge,

$$\begin{gathered} q_2=\text{2 fC} \\ =\left(\text{2 fC}\right)\left(\frac{{10}^{-15}\text{C}}{\text{1.0 fC}}\right) \\ =2\times {10}^{-15}\text{C} \end{gathered}$$

The charge,

role="math" localid="1662443184850" $$\begin{gathered} q_3=\text{3 fC} \\ =\left(\text{3 fC}\right)\left(\frac{{10}^{-15}\text{C}}{\text{1.0 fC}}\right) \\ =3\times {10}^{-15}\text{C} \end{gathered}$$

Convert the units of distance from $\text{cm}$to $\text{m}$.

role="math" localid="1662443206369" $$\begin{gathered} d=\text{2.54cm} \\ =\left(\text{2.54cm}\right)\left(\frac{\text{1.0m}}{\text{100cm}}\right) \\ =0.0254\text{m} \end{gathered}$$

Potential at a point due to a charge q is given as,

$\mathbf{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\frac{\mathbf{q}}{\mathbf{r}}$

Here, $\mathbf{r}$ is the distance of the point from the charge and$\frac{\mathbf{1}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}$is the Coulomb constant.

Schematic diagram of the system of charges due to which the charge has to be identified is drawn below.

The total potential at the point $\mathbf{P}$is the sum of the potentials due to each charge.

Formula:

The electric potential is define by,

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_o}·\frac{q}{R} \\ =\frac{kq}{R} \end{gathered}$$

Where, $V$is potential energy, $R$ is distance between the point charges, $q$is charge, and $k$ is the Coulomb’s constant having a value $9\times {10}^{9}N{m}^2/C^2$.

The total potential at point $P$due to given system of charges is,

$$\begin{gathered} V=\frac{1}{4\pi {\epsilon }_0}\left[\frac{q_1}{\sqrt{d^2+{\left(\frac{d}{2}\right)}^2}}-\frac{q_2}{\frac{d}{2}}-\frac{q_3}{\sqrt{d^2+{\left(\frac{d}{2}\right)}^2}}+\frac{q_3}{\sqrt{d^2+{\left(\frac{d}{2}\right)}^2}}-\frac{q_2}{\frac{d}{2}}+\frac{q_1}{\sqrt{d^2+{\left(\frac{d}{2}\right)}^2}}\right] \\ =\frac{1}{4\pi {\epsilon }_0}\left[\frac{2q_1}{\sqrt{d^2+{\left(\frac{d}{2}\right)}^2}}-\frac{4q_2}{d}\right] \end{gathered}$$

Substitute known values in the above equation.

$V=\left(9\times {10}^{9}N{m}^2/C^2\right)$$\left[\frac{2\left(5\times {10}^{-15}\text{C}\right)}{\sqrt{{\left(0.0254\text{m}\right)}^2+{\left(\frac{0.0254\text{m}}{2}\right)}^2}}-\frac{4\left(2\times {10}^{-15}\text{C}\right)}{0.0254\text{m}}\right]$

role="math" localid="1662444909542" $=\left(9\times {10}^{9}N{m}^2/C^2\right)$$\left[\frac{\left(10\times {10}^{-15}\text{C}\right)}{\sqrt{\text{0.00064516}+0.\text{00016129}}\text{m}}-\frac{\left(8\times {10}^{-15}\text{C}\right)}{0.0254}\right]$

$=\left(9\times {10}^{9}N{m}^2/C^2\right)$$\left[\frac{\left(10\times {10}^{-15}C\right)}{0.0284m}-314.961\times {10}^{-15}C/m\right]$

$=\left(9\times {10}^{9}N{m}^2/C^2\right)\left[352.113\times {10}^{-15}C/m-314.961\times {10}^{-15}C/m\right]$

$V=\left(9\times {10}^{9}N{m}^2/C^2\right)\left[37.152\times {10}^{-15}C/m\right]$

$=334.368\times {10}^{-6}V$

Hence, the total potential at the point $P$is $3.34\times {10}^{-4}\text{V}$.