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three equal point charges, each with charge 1.95 μc , are placed at the vertices of an equilateral triangle whose sides are of length 0.450 m . what is the electric potential energy u of the system? (take as zero the potential energy of the three charges when they are infinitely far apart.)



Answer :

The electric potential energy U of the system is 11.39 * [tex]10^{-2}[/tex] J

We know that,

U = [tex]q^{2}[/tex] / 4 π ε₀ d

where,

U = Electric potential energy

q = Point charge

ε₀ = Permittivity of free space

d = Length

Given that,

q = 1.95 * [tex]10^{-6}[/tex] C

d = 0.45 m

ε₀ = 8.85 * [tex]10^{-12}[/tex] [tex]C^{2}[/tex] N [tex]m^{2}[/tex]

U = [tex]U_{1}[/tex] + [tex]U_{2}[/tex] + [tex]U_{3}[/tex]

U = 3  [tex]q^{2}[/tex] / 4 π ε₀ d

U = 8.99 * [tex]10^{9}[/tex] * 3 * ( 1.3 * [tex]10^{-6})^{2}[/tex] / 0.45

U = 11.39 * [tex]10^{-2}[/tex] J

Electric potential energy is the amount of energy that is needed to move a charge against an electric field.

Therefore, the electric potential energy U of the system is 11.39 * [tex]10^{-2}[/tex] J

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