For two point charges, the sum in Equation 25.12 gives the following. In this ex

Question

For two point charges, the sum in Equation 25.12 gives the following.

In this example, q1 = 2.39 µC, r1 = 4.00 m, q2 = -6.26 µC, and r 2 = 5.00 m. Therefore, Vp reduces to the following.

W = U = q3V = q3(Vp - 0) = (2.76 10-6 C)(Vp) =  J

The negative sign is because the 2.76 µC charge is attracted to the combination of q1 and q2, which has a net negative charge. The 2.76 µC charge would naturally move toward the other charges if released from infinity, so the external agent does not have to do anything to cause them to move together. To keep the charge from accelerating, however, the agent must apply a force away from point P. Therefore, the force exerted by the agent is opposite the displacement of the charge, leading to a negative value of the work. Positive work would have to be done by an external agent to remove the charge from P back to infinity.

You can explore the value of the electric potential at point P and the electric potential energy of the system in Figure 25.12b using this Interactive Example.

Find the total potential energy of the system of three charges in the configuration shown in Fig. 25.12b in the example.

___________ J

Explanation / Answer

Yes the explanation and method are appropriate. The numerical values of the answers are (a) -5.88 x 10^3 Volt and (b) -1.614 x 10^(-2) Joule.

The total energy of the system is k(q1q2/r12 + q1q3/r13 + q2q3/r23) = -6.102 x 10^(-2) Joule.

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