Question

A point charge q₀ that has a charge of 0.5 μC is at the origin.

(a) Calculate the potential at x = 0.70 m (taking it to be zero at infinite distance).
V

(b) A second particle q that has a charge of 1 μC and a mass of 0.08 g is placed at x = 0.70 m. What is the potential energy of this system of charges?
J

(c) If the particle with charge q is released from rest, what will its speed be when it reaches x = 3 m?
m/s

Answer 1

a)

V = k*q/r

= 9*10^9*(0.5*10^-6)/0.7

= 6.43*10^3 V

b)

Potential energy, U =V*q'

= 6.53*10^3*(1*10^-6)

= 6.53*10^-3 J

c)

Change in PE = Change in KE

So, k*q*q'*(1/r1 - 1/r2) = 0.5*mv^2

So, 9*10^9*(0.5*10^-6)*(1*10^-6)*(1/0.7 - 1/3)= 0.5*8*10^-5*v^2

So, v = 11.1 m/s <-------- answer