Question

Determine the electrostatic potential energy of a system consisting of two charges $7\mu C$ and
$- 2\mu C$ (And with no external field) placed at $\left( { - 9cm,0,0} \right)$ and $\left( {9cm,0,0} \right)$ respectively.

Answer 1

To solve this question we first have to find the distance between the charges due to which electrostatic potential will come into picture. Afterward keep in mind that we have to change the unit of cm into m according to the demand of the SI unit of electrostatic potential. Then, we just have to put the values of variables in the equation of electrostatic potential.

Complete step by step solution:
Given that there is a system which consists of two charges $7\mu C$ and $- 2\mu C$ with no external field. Both the charges are being placed at $( - 9cm,0,0)$ and $(9cm,0,0)$ . And we have to find the electrostatic potential energy of the system.
For that we know the formula of electrostatic potential which is denoted by $U$ and give as $\Rightarrow U = \dfrac{1}{{4\pi { \in _ \circ }}} \times \dfrac{{{q_1}{q_2}}}{r}$
Here, $U$ is the electrostatic potential
${q_1},{q_2}$ are the two charges which are kept in the system
$r$ is the distance between the charges between ${q_1}$ and ${q_2}$
In the above case the point where the charges are kept is given as coordinate. So, for finding the distance we have to use the formula of finding the distance between the two points which is given by
$\Rightarrow D = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}}$
Now, by putting the values of coordinate in the above equation we will get the value of distance $D$ which will be equal to $r$
$\Rightarrow {x_1} = - 9cm,\,\,{x_2} = 9cm,\,\,{y_1} = 0,\,\,{y_2} = 0,\,{z_1} = 0,\,\,{z_2} = 0$
Then, the equation will become
$\Rightarrow D = \sqrt {{{(18)}^2} + {{(0)}^2} + {{(0)}^2}}$
And on solving it we get
$\Rightarrow D = 18cm$
So, we have $r = 0.18m$ as we had converted $cm$ into $m$ according to the electrostatic potential SI unit.
Then, putting the values of ${q_1},{q_2}$ and $r$ in the equation of electrostatic potential.
Now, our equation of electrostatic potential will become
$\Rightarrow U = \left( {9 \times {{10}^9}} \right) \times \left( {7 \times \dfrac{{ - 2}}{{0.18}}} \right)$ , where $\left( {1/4\pi { \in _ \circ }} \right) = 9 \times {10^9}$
After solving the equation we will get the value of $U = - 0.7J$

Therefore, the electrostatic potential of the system will be $- 0.7J$ .

Note: Remember the formula of electrostatic potential which is given by $U = \dfrac{1}{{4\pi { \in _ \circ }}} \times \dfrac{{{q_1}{q_2}}}{r}$
. And with that we have to find the distance between the two charges if they are given in a coordinate system. Although, these kinds of questions are easy but sometimes one may get confused, so try to approach the above process and go step by step to prevent the unnecessary confusions.