Question
How much work has to be done in assembling three charged particles at the vertices of an equilateral triangle as shown in figure?
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Answer
Amount of work done is assembling the charges is equal to the net potential energy
So, P.E. $=\text{U}_{12}+\text{U}_{13}+\text{U}_{23}$
$=\frac{\text{Kq}_1\text{q}_2}{\text{r}_{12}}+\frac{\text{Kq}_1\text{q}_3}{\text{r}_{13}}+\frac{\text{Kq}_2\text{q}_3}{\text{r}_{23}}$
$=\frac{\text{K}+10^{10}}{\text{r}}[4\times2+4\times3+3\times2]$
$=\frac{9\times10^9\times10^{10}}{10^{-1}}(8+12+6)$
$=9\times26$
$=234\text{J}$
So, P.E. $=\text{U}_{12}+\text{U}_{13}+\text{U}_{23}$
$=\frac{\text{Kq}_1\text{q}_2}{\text{r}_{12}}+\frac{\text{Kq}_1\text{q}_3}{\text{r}_{13}}+\frac{\text{Kq}_2\text{q}_3}{\text{r}_{23}}$
$=\frac{\text{K}+10^{10}}{\text{r}}[4\times2+4\times3+3\times2]$
$=\frac{9\times10^9\times10^{10}}{10^{-1}}(8+12+6)$
$=9\times26$
$=234\text{J}$
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