Three charges $+ 1 μ C$ , $+ 3 μ C$ and $- 5 μ C$ are kept at the vertices of an equilateral triangle of sides $60 c m$ . Find the electrostatic potential energy of the system of charges.

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Solution

Given data:
$q_{A} = + 1 \times 10^{- 6} C$ ; $q_{B} = + 3 \times 10^{- 6} C$ ; $q_{C} = - 5 \times 10^{- 6} C$
$r = 60 c m$ (or) $6 m$
Solution:
Electrostatic potential energy $=$
$U = \frac{1}{4 π ε_{0}} [\frac{q_{A} q_{B}}{A B} + \frac{q_{B} q_{C}}{B C} + \frac{q_{A} q_{C}}{A C}]$
$= \frac{9 \times 10^{9}}{0.6} \times 10^{- 12} [3 - 15 - 5]$
$V = \frac{- 9 \times 10^{9} \times 10^{- 12} \times 17}{0.6} = V = - 0.2553$

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Three charges +1μC, +3μC and −5μC are kept at the vertices of an equilateral triangle of sides 60cm. Find the electrostatic potential energy of the system of charges.

Given data:qA=+1×10−6C; qB=+3×10−6C; qC=−5×10−6Cr=60cm (or) 6mSolution:Electrostatic potential energy =U=14πε0[qAqBAB+qBqCBC+qAqCAC] =9×1090.6×10−12[3−15−5]V=−9×109×10−12×170.6=V=−0.2553

Three charges $+ 1 μ C$ , $+ 3 μ C$ and $- 5 μ C$ are kept at the vertices of an equilateral triangle of sides $60 c m$ . Find the electrostatic potential energy of the system of charges.