Question

A positively charged particle Q1=35 nC is held fixed at the origin A negatively charged particle O2=24 nC of mass m = 8.5 ug is located a distance d = 45cm from the positively charged particle along the positive x-axis.

Part (a) What is the magnitude of the electric force, F in newtons, that acts on the charge Q2? 

(b) What is the direction of the force on charge Q2?

Part (c) what is the magnitude of the acceleration, a in m/s², experienced by charge Q2? 

Answer 1

A.)

12 = kq1q2/d^2

=) F12 = 9×10^9 × 35×10^-6 × 24×10^-6/(0.45)^2

=) F12 = 37.34N

B.) Direction of force will be attractive ,that means along left towards -ve x axis.

C.) Acceleration a = F12/m = 37.34/8.5×10^-6

=) a = 4.39×10^6m/s^2