Question

answer all blank boxes please

Strategy Use Equations 25.12. Solution A For two point charges, the sum in Equation 25.12 gives the following. V;=k + In this example, 91 - 2.50 C, -4.00 m, 42 = -6.05 pc, and I 2 = 5.00 m. Therefore, V, reduces to the following V = (8.99 x 10 Nm?/C) x (230 x 10°C -6.05 x 10°C 100 m 5.00 m The work done is equal to the change in the potential energy given by the following. W = AU - AV = 3(Vp - 0) = (3.15 x 106 CV) - The negative sign is because the 3.15 C charge is attracted to the combination of , and 42, which has a net negative charge. The 3.15 uC 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 25.12 using this Interactive Example Exercise 25.3 Hints: Getting Started I'm stuck Find the total potential energy of the system of the shown in Fig. 25.12b in the example changes in the configuration 2 / / a w

Answer 1

up = ke (a ta vp={8.99 X10 N-~*%*)+(2,5X10€ + (-6.35XPO). Vec - 5259.15 v um sm ③ W = su-qov = 23 Cup-o) = (3.15 8100c (up) = -0.01656 J figure doing 25.12b enercise is missing 25.3 -