Question

Four identical charges (+3.0 \mu C each)are brought from infinity and fixed to a straight line. The chargesare located 0.50 m apart. Determinethe electric potential energy of this group.

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asked 2020-12-28
Four identical charges (+3.0 \(\displaystyle\mu{C}\) each)are brought from infinity and fixed to a straight line. The chargesare located 0.50 m apart. Determinethe electric potential energy of this group.

Answers (1)

2020-12-29

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Initially there is no charge in the space. Whenthe charge \(\displaystyle{Q}_{{1}}\) is brought from infinity to a point inspace, the potential energy is zero.
Secondly when the charge \(\displaystyle{Q}_{{2}}\) is broughtfrom infinity and placed at a distance say \(\displaystyle{r}_{{{12}}}\)
from thecharge \(\displaystyle{Q}_{{1}}\), the P.E. of the system is \(E_1=\frac{Q_1Q_2}{4\pi\epsilon_or_{12}^2}\)
\(\displaystyle{\frac{{{\left({3}\times{10}^{{-{6}}}\right)}^{{2}}\times{9}\times{10}^{{9}}}}{{{\left({0.5}\right)}^{{2}}}}}={0.324}\ {J}\)
When the third charge is brought to a point which is at a distance 0.5 m from the charge \(\displaystyle{Q}_{{2}}\) and 1.0m from the charge \(\displaystyle{Q}_{{2}}\) and 1.0 m from \(\displaystyle{Q}_{{1}}\), this has to be taken in a potential; due to the above two charges. Hence the potential energy of \(\displaystyle{Q}_{{3}}\) due to these two charges is
\(E_2=Q_3[\frac{Q_1}{4\pi\epsilon_0r_{13}^2}+\frac{Q_2}{4\pi\epsilon_or_{23}^2}]=\frac{(3\times10^{-6})^2\times9\times10^9}{0.5^2}+\frac{(3\times10^{-6})\times9\times10^9}{1.0^2}\)
So the total potential energy, \(\displaystyle={E}_{{1}}+{E}_{{2}}+{E}_{{3}}={1.17}\ {J}\)

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