The point charges $- 5 n C, 10 n C$ and $12 n C$ are located at $(0, 0, 0), (0, 0, 1)$ and $(0, 0, 2)$ respectively. Find the total energy of the system.

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Solution

$\begin{matrix} q_{1} = - 5 \times 10^{- 9} C q_{2} = 10 \times 10^{- 9} C q_{3} = 12 \times 10^{- 9} C r_{1} = (0, 0, 0) r_{2} = (0, 0, 1) r_{3} = (0, 0, 2) T . E . = \frac{1}{4 π ε_{o}} [\frac{q_{1} q_{2}}{| r_{2} - r_{1} |} + \frac{q_{3} q_{2}}{| r_{3} - r_{2} |} + \frac{q_{1} q_{3}}{| r_{3} - r_{1} |}] | r_{2} - r_{1} | = \sqrt{1^{2}} = 1 | r_{3} - r_{2} | = \sqrt{1^{2}} = 1 | r_{3} - r_{1} | = \sqrt{2^{2}} = 2 \end{matrix}$
substitute and calculating value:

$T . E . = \frac{1}{4 π ϵ_{o}} [- 50 \times 10^{- 18} + 120 \times 10^{- 18} + \frac{- 60 \times 10^{- 18}}{2}]$
$T . E . = \frac{10^{- 18}}{4 π ϵ_{o}} [40]$

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The point charges −5nC,10nC and 12nC are located at (0,0,0),(0,0,1) and (0,0,2) respectively. Find the total energy of the system.

q1=−5×10−9Cq2=10×10−9Cq3=12×10−9Cr1=(0,0,0)r2=(0,0,1)r3=(0,0,2)T.E.=14πεo[q1q2|r2−r1|+q3q2|r3−r2|+q1q3|r3−r1|]|r2−r1|=√12=1|r3−r2|=√12=1|r3−r1|=√22=2substitute and calculating value:T.E.=14πϵo[−50×10−18+120×10−18+−60×10−182]T.E.=10−184πϵo[40]

The point charges $- 5 n C, 10 n C$ and $12 n C$ are located at $(0, 0, 0), (0, 0, 1)$ and $(0, 0, 2)$ respectively. Find the total energy of the system.