Question

Two stationary point charges of 4.00nC and 2.00nC are separated by a distance of 60.0cm ....

Two stationary point charges of 4.00nC and 2.00nC are separated by a distance of 60.0cm . An electron is released from rest at a point midway between the charges and moves along the line connecting them. What is the electric potential energy of the electron when it is at the midpoint? In J.

14.0cm from the 4.00nC charge?

Please help!!! Thank you in advance!

Homework Answers

Answer #1


a)
At mid point
distance R = 30 cm =0.3 m
Q1 = 4nC =4x10-9 C
Q2 = 2 nC = 2 x10-9 C
charge of the electron Q=-1.60x10-19 C
electric potential energy when it is at mid point
U =kQ1Q/R + kQ2Q/R = kQ*[Q1+Q2]/R
U = 9x109*(-1.6x10-19)[4x10-9 + 2x10-9 ]/0.3 = 288x10-19 J = 2.88x10-17 J
b)
potential energy 14 cm from the 4nC charge, and 46 cm from the 2nC charge
U2=KQ1Q/0.14 + kQ2Q/0.46 =kQ[Q1/0.14+Q2/0.46]
U2= 9x10^9*(-1.6x10-19)[4x10-9/0.14 + 2x10-9/0.46]=474x10-19 J =4.74x10-17 J

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