Question

What is the electric potential at the point indicated with the dot

in FIGURE EX25.30?

Short answer

The value of electric potential energy at the point indicated with the dot is 1770VV

Step-by-step solution

Definition of  electric potential  energy  

The value of potential charge of the q point is V

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

Here q is the charge value , r is the distance ${ϵ}_0$Permittivity of the free space.

Then the total value of potential $\sum _i{V}_i=\frac{1}{4\pi {\epsilon }_0}\frac{q_i}{r_i}$

The magnitude of all charges in Point 1 and 3 is 2.0 nC.

$\begin{array}{rcl}{q}_1& =& q_3=\left(2.0nC\right)\left(\frac{{10}^{-19}nC}{1nC}\right)\\ & =& 2.0\times {10}^{-19}C\end{array}$

Magnitude of all charges is in Point 2 is 4.0 nC

$\begin{array}{rcl}Now{q}_2& =& \left(4.0nC\right)\left(\frac{{10}^{-19}nC}{1nC}\right)\\ & =& 4.0\times {10}^{-19}C\end{array}$

Point indication with the dot  points 

The value of distance from the unit positive charge of the 2

$\begin{array}{rcl}& & \\ r_2& =& \sqrt{\left(4^2\right)}+\left(3^2\right)\\ & =& 5cm\\ & & \\ V& =& \left(\frac{1}{4\pi {\epsilon }_0}\frac{q_1}{r_1}\right)+\left(\frac{1}{4\pi {\epsilon }_0}\frac{q_2}{r_2}\right)+\left(\frac{1}{4\pi {\epsilon }_0}\frac{q_3}{r_3}\right)\\ & & \\ \frac{1}{4\pi {\epsilon }_0}& =& 9\times {10}^{9}N.m^2\\ & & \\ & & \\ & & \end{array}$

Now substitutingall the value to this equation;

$\begin{array}{rcl}V& =& 9\times {10}^{9}N.m^2\times \left(\frac{2\times {10}^{-19}C}{3\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)}+\frac{4\times {10}^{-19}C}{5\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)}+\frac{2\times {10}^{-19}C}{4\left({\displaystyle \frac{{10}^{-2}m}{1cm}}\right)}\right)\\ & =& 1770V\end{array}$