Two equal positive charges each of value 'q' are placed at points A and B, where AB = 3x . A third charge -3q is placed at point C at a distance x from A on AB. The potential energy of the system is nearly ( ₀ = permittivity of free space)

Step-by-step solution

The potential energy of the system is the sum of the potential energies of all pairs of charges. U = U_ AB + U_ BC + U_ CA U = 1 4 ₀ ( q_A q_B r_ AB + q_B q_C r_ BC + q_C q_A r_ CA ) Given q_A = q , q_B = q , and q_C = -3q . The distances between the charges are: r_ AB = 3x r_ CA = x r_ BC = 3x - x = 2x Substituting these values into the potential energy expression: U = 1 4 ₀ ( q q 3x + q (-3q) 2x + (-3q) q x ) U = q^2 4 ₀ x ( 1 3 - 3 2 - 3 ) U = q^2 4 ₀ x ( 2 - 9 - 18 6 ) U = -25 6 ( q^2 4 ₀ x ) U -4.167 ( q^2 4 ₀