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Three equal 1.75-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.250 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.

Respuesta :

Answer:

[tex]U = 0.33 J[/tex]

Explanation:

As we know that electrostatic potential energy of a system of charge is given by equation

[tex]U = \frac{kq_1q_2}{r}[/tex]

here we know that

[tex]q_1 = q_2 = 1.75 \mu C[/tex]

now we also know that

r = distance between two charges = 0.250 m

now here three charges are placed at the vertices of the triangle

so here net electric potential energy of all three charges is 3 times the energy of one pair of charge

[tex]U = \frac{3kq_1q_2}{r}[/tex]

[tex]U = \frac{3(9\times 10^9)(1.75 \mu C)^2}{0.250}[/tex]

[tex]U = 0.33 J[/tex]

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