Question

A charged particle A exerts a force of 2.65 N to the right on charged particle B when the particles are 13.8 mm apart. Particle B moves straight away from A to make the distance between them 19.0 mm. What vector force does particle B then exert on A

Answer 1

-1.4N

Step-by-step explanation:

When the distance between A and B is 13.8mm (0.0138m), the force exerted by A on B is 2.65N:

F(A,B) = k*qA*qB/r²

=> 2.65 = k*qA*qB / 0.0138²

=> k*qA*qB = 2.65 * 0.0138²

k*qA*qB = 0.0005046

When the distance between them increases to 19mm (0.019m), the force exerted by A on B becomes:

F(A,B) = k*qA*qB / 0.019²

Given that k*qA*qB = 0.0005046,

F(A,B) = 0.0005046/0.019²

F(A,B) = 1.4N

The force exerted by B on A will be in the opposite direction of the force exerted by A on B, which means that the force exerted by B on A will be:

F(B, A) = -1.4N