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:
$P.E={\displaystyle \frac{k{q}_1{q}_2}{r}}$

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 $P.E={\displaystyle \frac{k{q}_1{q}_2}{r}}$ using the given we have
$⟹T.P.E={\displaystyle \frac{k\times 2\times {10}^{-8}\times 3\times {10}^{-8}}{1}}+{\displaystyle \frac{k\times -2\times {10}^{-8}\times 3\times {10}^{-8}}{1}}+{\displaystyle \frac{k\times -2\times {10}^{-8}\times 1\times {10}^{-8}}{1}}+{\displaystyle \frac{k\times 1\times {10}^{-8}\times 2\times {10}^{-8}}{1}}+{\displaystyle \frac{k\times 2\times {10}^{-8}\times -2\times {10}^{-8}}{\sqrt{2}}}+{\displaystyle \frac{k\times 1\times {10}^{-8}\times 3\times {10}^{-8}}{\sqrt{2}}}$
$⟹T.P.E={\displaystyle \frac{k\times 6\times {10}^{-16}}{1}}+{\displaystyle \frac{k\times -6\times {10}^{-16}}{1}}+{\displaystyle \frac{k\times -2\times {10}^{-16}}{1}}+{\displaystyle \frac{k\times 2\times {10}^{-16}}{1}}+{\displaystyle \frac{k\times -4\times {10}^{-16}}{\sqrt{2}}}+{\displaystyle \frac{k\times 3\times {10}^{-16}}{\sqrt{2}}}$on simplification, we get
$⟹T.P.E={\displaystyle \frac{k\times -4\times {10}^{-16}}{\sqrt{2}}}+{\displaystyle \frac{k\times 3\times {10}^{-16}}{\sqrt{2}}}$
$⟹T.P.E={\displaystyle \frac{k\times {10}^{-16}}{\sqrt{2}}}\times (3-4)$
On further simplification, we have
$⟹T.P.E={\displaystyle \frac{k\times -1\times {10}^{-16}}{\sqrt{2}}}$
Since $k=9\times {10}^9$, substituting, we have
$⟹T.P.E={\displaystyle \frac{9\times {10}^9\times -1\times {10}^{-16}}{\sqrt{2}}}={\displaystyle \frac{9\times {10}^{-7}}{\sqrt{2}}}$
$\therefore T.P.E={\displaystyle \frac{9\times {10}^{-7}}{\sqrt{2}}}J$
Thus the required total potential energy at the centre of the given square with side $a=1.0m$${\displaystyle \frac{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.