Electrostatics

Question

Two charges q1 = −3.20 nC and q2 = +8.15 nC are at a distance of 1.28 µm from each other. q1 is fixed at its location while q2 is released from rest.

(a) What is the kinetic energy of the charge q2 when it is 0.340 µm from q1?

 

(b) The charge q2 has a mass m2 = 7.85 µg. What is its speed when it is 0.340 µm from q1?

 

 

For part A, I thought I would have to use conservation of energy and to solve for final kinetic energy, but I'm confused how to go from there. If you could just steer me in the right direction with what equation(s) to use, I can take it from there. 

Answer 1

Find the difference in PE between the first and last position of moving charge. This value will be the KE of q2.

 

Equate this KE to 1/2( mV²)

 

Where m = 7.85x 10 ^-9 kg  

 

Speed of q2 = ( 2x change in PE (in J)/ 7.85x 10-6)^1/2  m/s

 

Note 1 n C = 10^-9 C

and 1 micro gm is 10^-6 gm = 10^-9 kg and 1 micro m = 10^-6 m

m2 = 7.85 x 10^-9 kg
Hi Megan... Hope it works.. if not get back to me