What is the total electrostatic potential ene | vedclass

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(A) The system consists of three charges $q_1 = q_2 = q_3 = 10 \ \mu C = 10 	imes 10^{-6} \ C$ placed at the vertices of an equilateral triangle with

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(A) The system consists of three charges $q_1 = q_2 = q_3 = 10 \ \mu C = 10 	imes 10^{-6} \ C$ placed at the vertices of an equilateral triangle with side length $r = 10 \ cm = 0.1 \ m$.The total electrostatic potential energy $U$ of a system of point charges is given by the sum of potential energies of all pairs:$U = \frac{1}{4\pi \varepsilon_0} \left( \frac{q_1 q_2}{r_{12}} + \frac{q_2 q_3}{r_{23}} + \frac{q_3 q_1}{r_{31}} ight)$Since all charges and distances are equal:$U = 3 	imes \left( \frac{1}{4\pi \varepsilon_0} \frac{q^2}{r} ight)$Substituting the values:$U = 3 	imes 9 	imes 10^9 	imes \frac{(10 	imes 10^{-6})^2}{0.1}$$U = 27 	imes 10^9 	imes \frac{100 	imes 10^{-12}}{0.1}$$U = 27 	imes 10^9 	imes 1000 	imes 10^{-12}$$U = 27 	imes 10^{12} 	imes 10^{-12} = 27 \ J$

What is the total electrostatic potential energy of the given system in $J$? (Given: $\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9} \ N \cdot m^2/C^2$)

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