A point charge Q = +4.60 C is held fixed at the origin. A second point charge q

Question

A point charge Q = +4.60 C is held fixed at the origin. A second point charge q = +1.20 C with mass of 2.80 x 10^-4 kg is placed on the x axis, 0.250 m from the origin. (a) What is the electric potential energy U of the pair of charges? (Take U to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) 0.500 m; (ii) 5.00 m; (iii) 50.0 m?

Please show all equations used and steps to obtain the answers for all the points.

Thanks!

Explanation / Answer

Electric potential is

U = k Qq/r^2

where Q is the point charge, q is the charge of the mass with charge, k is coulombs constant, and are is the separation between them.

b.) From conservation of Energy, we know that for any given system, any kinetic energy gained is equal to the loss of potential energy (PE) of that system.

That is, KE1 +PE1 = KE2 +PE2

Kinetic energy is defined as

KE = (1/2) m v^2

using the conservation of momentum equation, setting KE1 = 0 you get

PE1-PE2 = KE2. = change in potential energy

Since potential energy changes only when distance from charge changes, you can set your PE1 distance (r) to 0.250 m, and PE2 distance to be any distance from origin given in part b

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