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Two spherical point charges each carrying a charge of 40 C are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1, what is the length of the spring when the charges are in equilibrium?

Respuesta :

Answer:

[tex]L = 20 + 37 = 57 cm[/tex]

Explanation:

As we know that two charges connected with spring is at equilibrium

so here force due to repulsion between two charges is counter balanced by the spring force between them

so here we have

[tex]F_e = F_{spring}[/tex]

here we have

[tex]\frac{kq_1q_2}{r^2} = kx[/tex]

[tex]\frac{(9 \times 10^9)(40 \mu C)(40 \mu C)}{(0.20 + x)^2} = 120 x[/tex]

[tex]14.4 = (0.20 + x)^2 ( 120 x)[/tex]

by solving above equation we have

[tex]x = 0.37 m[/tex]

so the distance between two charges is

[tex]L = 20 + 37 = 57 cm[/tex]

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