A charge of +3.00 µC is placed at the point (0 m, 0 m, 0 m); a charge of -4.50 µ

Question

A charge of +3.00 µC is placed at the point (0 m, 0 m, 0 m); a charge of -4.50 µC is placed at the point (2.00 m, -5.00 m, 7.00 m); finally, a charge of -1.00 µC is placed at the point (1.00 m, 1.00 m, 1.00 m). (a) What is the electric potential energy of this configuration? (b) If a neutron (a particle with the mass of the proton but no electric charge) is brought in from infinity and placed at the point (-3.00 m, 2.00 m, 5.50 m), how does the electric potential energy of the configuration change?

Explanation / Answer

q1 = +3.00 µC at (0 m, 0 m, 0 m);

q2 = -4.50 µC at (2.00 m, -5.00 m, 7.00 m);

q3 = -1.00 µC at (1.00 m, 1.00 m, 1.00 m).

distance between q1 and q2 is r12 = (2.002 + 5.002 + 7.002) = 8.83 m

distance between q1 and q3 is r12 = (1.002 + 1.002 + 1.002) = 1.73 m

distance between q2 and q3 is r23 = (1.002 + 6.002 + 6.002) = 8.54 m

(a) What is the electric potential energy of this configuration?

U = kq1q2/r12 + kq1q3/r13 + kq2q3/r23 = -0.0246 J

b) no change, because a neutron has no charge, and won't affect electric potential energy.

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