Need exercise to 25.3 Electric Potential Due to Two Point Charges Problem A q_1

Question

Need exercise to 25.3

Electric Potential Due to Two Point Charges Problem A q_1 = 1.7 1 mu C point charge is located at the origin, and a second point charge of q_2 = -6.12 mu C is located on the y axis at the position (0, 3.00) m as in Figure 25.12a. Find the total electric potential due to these charges at point P, whose coordinates are (4.00, 0) m. How much work is required to bring a q_3 = 3.47 mu C point charge from infinity to the point P (as shown in Fig. 25.12b)? Strategy Use Equations 25.12. For two point charges, the sum in Equation 25.12 gives the following. in this example, q_1 = 1.7 1 mu C, r_1 = 4.00 m, q_2 = -6.12 mu C, and r_2 = 5.00 m. Therefore, V_p reduces to the following. V_p = (8.99. times 10^9 N middot m^2/C^2) x (1.71 times 10^-6 C/4.00 m + -6.12 times 10^-6 C/5.00 m) =-7160 The work done is equal to the change in the potential energy given by the following. W = Delta U = q_3DeltaV = q_3 (V_p - 0) = (3.47 times 10^-6 C) (V_p) = -0.0248 The negative sign is because the 3.47 mu C charge is attracted to the combination of q_1 and q_2. which has a net negative charge. The 3.47 mu C 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. -16.217E-3 How is potential energy defined? Determine the potential energy for each combination of two charges and simply add them together.

Explanation / Answer

For a system of 3 charges, the total PE = the sum of the PEs of each possible pair of charges.

Thus PE = k(q1q2/r12 + q1q3/r13 + q2q3/r23)

It's probably obvious, but since r12 = 3 m and r13 = 4 m, r23 = 5 m

q1 = 1.71 uC

q2 = -6.12 uC

q3 = 3.47 uC

PE = (9 x 10^9)[(1.71 x -6.12 x 10^-12) / 3 + (1.71 x 3.47 x 10^-12) / 4 + (-6.12 x 3.47 x 10^-12) / 5 ]

PE = (9 x 10^9 x 10^-12)[-3.488 + 1.4834 - 4.2473]

PE = -5.627 x 10^-2 J

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