A point charge q2 = -1.4 ?C is fixed at the origin of a co-ordinate system as sh

Question

A point charge q2 = -1.4 ?C is fixed at the origin of a co-ordinate system as shown. Another point charge q1 = 2.3 ?C is is initially located at point P, a distance d1 = 7.4 cm from the origin along the x-axis. 1) What is ?PE, the change in potenial energy of charge q1 when it is moved from point P to point R, located a distance d2 = 3.0 cm from the origin along the x-axis as shown? 2) The charge q2 is now replaced by two charges q3 and q4 which each have a magnitude of -0.7 ?C, half of that of q2. The charges are located a distance a = 1.8 cm from the origin along the y-axis as shown. What is ?PE, the change in potential energy now if charge q1 is moved from point P to point R?

Explanation / Answer

a) The initial potential energy is

(U_i = kq_1q_2/d_1 = (8.99*10^9Nm^2/C^2)(2.3*10^{-6}C)(-1.4*10^{-6}C)/(0.074m))

--> (U_i = -0.391J)

The final potential enegy is

(U_f = kq_1q_2/d_2 = (8.99*10^9Nm^2/C^2)(2.3*10^{-6}C)(-1.4*10^{-6}C)/(0.03m))

--> (U_f = -0.965J)

So,

(Delta U = U_f - U_i = -0.965J + 0.391J = -0.574J)

b) The fact that I can't see whether q3 is above or below the x axis (the same goes for q4) is actually irrelevant. Position on the graph doesn't matter, only distance between charges. Since both q3 and q4 have the same charge, and are the same distance from q1, the potential energy due to each charge will be the same. At point P, the potential energy is

(U_i = 2kq_1q_3/d_1 = 2(8.99*10^9Nm^2/C^2)(2.3*10^{-6}C)(-1.4*10^{-6}C)/sqrt{(0.074m)^2 + (0.018m)^2})

--> (U_i = -0.760J)

And the final potential energy is

(U_f = 2kq_1q_3/d_2 = 2(8.99*10^9Nm^2/C^2)(2.3*10^{-6}C)(-1.4*10^{-6}C)sqrt{(0.03m)^2 + (0.018m)^2})

--> (U_f = -1.654J)

So,

(Delta U = U_f - U_i = -0.894J)

Hope this helps

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