Question

1- How far from a -7.80 μC point charge must a 2.10 μC point charge be placed in order for the electric potential energy of the pair of charges to be -0.500 J ? (Take the energy to be zero when the charges are infinitely far apart.)

d= ------m

2- Two stationary positive point charges, charge 1 of magnitude 3.50 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 37.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

Express your answer in meters per second.

Answer 1

a) E = k (q1) (q2) / r

500 = (9x 10^9 ) (-7.80 x 10^-6 ) (2.10 x 10^-6) /r

r = 0.29484 x 10^ -3 m

b) we will use conservation of energy,

initial PE = final PE + KE

Initial pE =( 9 x10^9 x 1.6 x 10^-19) / (0.185 ) ( 3.50 x 10^-9 + 2 x 10^-9) = ( 77.837 x 10^ -19 ) ( 5.50) = 428. 108 x 10^-19 J

Final PE = (  9 x10^9 x 1.6 x 10^-19) ) ( 3.50 x 10^-9 / 0.1+ 2 x 10^-9/0.27) = 610.67 x 10^ -19 J

KE = 182.562 x 10^ -19 J

1/2 mv^2 = 182.562 x 10^ -19 J

0.5 ( 9.109 x 10^-31 ) v^2 = 182.562 x 10^ -19

v^2= 40.08 x 10^ 12

v = 6.33 x 10^6 m/s