Question

The electric potential at the dot in FIGURE EX25.32 is 3140 V. What is charge q ?

Short answer

The value of the potential enegy of the dot points is 10nC

Step-by-step solution

 Explanation of  the  the potential enegy  of a point

The equation of the potential energy of a dot point is

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

Where q is the charge point, r is the distance and ${\epsilon }_0$is the permittivity free space.

the value of the distance d from the q point of the figure to A point is

$\begin{array}{rcl}d& =& \sqrt{{\left(2.0cm\right)}^2+{\left(4.0cm\right)}^2}\\ & =& 4.47cm\end{array}$

$\begin{array}{rcl}{r}_1& =& 2.0cm\\ & =& 0.02m\\ r_2& =& 4.0cm\\ & =& 0.04m\\ d& =& 4.47cm\\ & =& 0.0447m\end{array}$

The potential energy at the point A for the responsible charge $q_1$

$\begin{array}{rcl}{V}_1& =& \frac{1}{\left(4\pi {\epsilon }_0\right)}\frac{q_1}{r_1}\\ \frac{1}{\left(4\pi {\epsilon }_0\right)}& =& 9.0\times {10}^{9}N{m}^2/C^2\\ q_1& =& 5.0\times {10}^{-9}C\\ r_1& =& 0.02m\\ V_1& =& 9.0\times {10}^{9}N{m}^2/C^2\times \frac{5.0\times {10}^{-9}C}{0.02m}\\ & =& 2250V\\ & & \end{array}$

The theory of the total  energy 

The potential energy at the point A for the responsible charge$q_2$

$\begin{array}{rcl}{V}_2& =& \frac{1}{\left(4\pi {\epsilon }_0\right)}\frac{q_2}{r_2}\\ \frac{1}{\left(4\pi {\epsilon }_0\right)}& =& 9.0\times {10}^{9}N{m}^2/C^2\\ q_2& =& -5.0\times {10}^{-9}C\\ r_2& =& 0.04m\\ V_2& =& 9.0\times {10}^{9}N{m}^2/C^2\frac{(-5.0\times {10}^{-9}C)}{0.04m}\\ & =& 1125V\\ & & \end{array}$

The potential energy at the point A for the responsible charge q is

$\begin{array}{rcl}{V}_3& =& \frac{1}{\left(4\pi {\epsilon }_0\right)}\frac{q_{}}{d}\\ \frac{1}{\left(4\pi {\epsilon }_0\right)}& =& 9.0\times {10}^{9}N{m}^2/C^2\\ d& =& 0.0447m\\ V_3& =& 9.0\times {10}^{9}N{m}^2/C^2\frac{\left(q\right)}{0.0447m}\\ & =& (201\times {10}^{9}V/C)q\end{array}$

$\begin{array}{rcl}{V}_3& =& \frac{1}{\left(4\pi {\epsilon }_0\right)}\frac{q_{}}{d}\\ \frac{1}{\left(4\pi {\epsilon }_0\right)}& =& 9.0\times {10}^{9}N{m}^2/C^2\\ d& =& 0.0447m\\ V_3& =& 9.0\times {10}^{9}N{m}^2/C^2\frac{\left(q\right)}{0.0447m}\\ & =& (201\times {10}^{9}V/C)q\end{array}$

The total potential energy
localid="1648300123891" $\begin{array}{rcl}& & V_1+V_2+V_3\\ & =& 2250V+1125V+(201\times {10}^{9}V/C)q\end{array}$

As the value of V has been said 3140V

localid="1648300321471" $\begin{array}{rcl}{V}_T& =& V_1+V_2+V_3\\ & =& 2250V+1125V+\left[(201\times {10}^{9}V/C)\times q\right]+3140V\\ & =& 2250V+1125V+\left[(201\times {10}^{9}V/C)\times q\right]\\ q& =& \frac{2015V}{(201\times {10}^{9}V/C)}\\ & =& 10nC\\ & & \end{array}$

Therefore, the value of the potential energy of the dot points is 10nC.