Question

What is the electric potential energy of the charge configuration as shown in the figure? Assume that
 $q_1=+1.0\times 10^{-8}C$, $q_2=-2.0\times 10^{-8}C$, $q_3=+3.0\times 10^{-8}C$, $q_4=+2.0\times 10^{-8}C$and $a=1.0m$

Answer 1

Hint: We know that the potential difference between two points in any electrical circuit is defined as the energy required to move a unit positive charge between the two points and potential energy is the energy stored by the system to do the same. Here, using the formula for the potential energy , we can find the net potential energy at the centre of the square as discussed below.

Formula used:
$P.E=\dfrac{kq_1q_2}{r}$

Complete step by step solution:
Given that, $q_1=+1.0\times 10^{-8}C$, $q_2=-2.0\times 10^{-8}C$, $q_3=+3.0\times 10^{-8}C$, $q_4=+2.0\times 10^{-8}C$and $a=1.0m$, then the centre of the square will be $O$ and the distance of the diagonals will be $\sqrt 2$, as shown in the figure below

Then the potential difference at the O due to the charges will be the sum of potential energies at the centre which is given as
$T.P.E=P.E_{AB}+P.E_{BC}+P.E_{CD}+P.E_{DA}+P.E_{AC}+P.E_{DB}$
Substituting, for $P.E=\dfrac{kq_1q_2}{r}$ using the given we have
$\implies T.P.E=\dfrac{k\times 2\times 10^{-8}\times 3\times 10^{-8}}{1}+\dfrac{k\times -2\times 10^{-8}\times 3\times 10^{-8}}{1}+\dfrac{k\times -2\times 10^{-8}\times 1\times 10^{-8}}{1}+\dfrac{k\times 1\times 10^{-8}\times 2\times 10^{-8}}{1}+\dfrac{k\times 2\times 10^{-8}\times -2\times 10^{-8}}{\sqrt 2}+\dfrac{k\times 1\times 10^{-8}\times 3\times 10^{-8}}{\sqrt 2}$
$\implies T.P.E=\dfrac{k\times 6\times 10^{-16}}{1}+\dfrac{k\times -6\times 10^{-16}}{1}+\dfrac{k\times -2\times 10^{-16}}{1}+\dfrac{k\times 2\times 10^{-16}}{1}+\dfrac{k\times -4\times 10^{-16}}{\sqrt 2}+\dfrac{k\times 3\times 10^{-16}}{\sqrt 2}$on simplification, we get
$\implies T.P.E=\dfrac{k\times -4\times 10^{-16}}{\sqrt 2}+\dfrac{k\times 3\times 10^{-16}}{\sqrt 2}$
$\implies T.P.E=\dfrac{k\times 10^{-16}}{\sqrt 2}\times (3-4)$
On further simplification, we have
$\implies T.P.E=\dfrac{k\times -1\times 10^{-16}}{\sqrt 2}$
Since $k=9\times 10^9$, substituting, we have
$\implies T.P.E=\dfrac{9\times 10^{9}\times -1\times 10^{-16}}{\sqrt 2} =\dfrac{9\times 10^{-7}}{\sqrt 2}$
$\therefore T.P.E =\dfrac{9\times 10^{-7}}{\sqrt 2}J$
Thus the required total potential energy at the centre of the given square with side $a=1.0m$$\dfrac{9\times 10^{-7}}{\sqrt 2}J$
Additional Information:
We also know that electric current is produced due to motion of charges. The two are related by Ohm's law.

Note: If the unit positive charge is moved from a region of high potential to a region of low potential, then the energy is emitted during the process, or work is done by the system. Similarly, if the unit positive charge is moved from a region of low potential to high potential, then energy is absorbed, or work is done on the system.