Two particles, with charges of 50.0 nC and -50.0 nC, are placed at the points wi

Question

Two particles, with charges of 50.0 nC and -50.0 nC, are placed at the points with coordinates (0, 8.00) cm and (0, -8.00) cm as shown below. A particle with charge 25.0 nC is located at the origin.

Two particles, with charges of 50.0 nC and -50.0 nC, are placed at the points with coordinates (0, 8.00) cm and (0, -8.00) cm as shown below. A particle with charge 25.0 nC is located at the origin. Two particles, with charges of 50.0 nC and -50.0 n10-13kg and a charge of100.0nC, is released from rest at the point (6.00, 0) cm. Find its speed after it has moved freely to a very large distance away. (a) Find the electric potential energy of the configuration of the three fixed charges. J (b) A fourth particle, with a mass of 1.68

Explanation / Answer

Alright, so the potential of a configuration is the sum of the potentials between each charge. The expression for each charge is V=kq1q2/r. So, naming the charges from top to bottom q1, q2, q3, and let's use q=25 nc, r=8 cm. The expression for total potential is:

(2kq^2/r)+(-2kq^2/r)+(-4kq^2/2r)

Simplifying,

-2kq^2/r

Well, that's nice, 2 of those expressions cancel each other out. Plugging in numbers (and remember to use the right units):

-2(8.8976E9 m/F)*(25E-9 C)^2/(8E-2 m) = -1.38E-4 J

Or, -139 microJoules. (note: I use E to mean x10^. It's a scientific notation thing)

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