Question

Two identical point charges in outer space are held apart at a distance D. As soon as the charges are released, each begins moving at an acceleration a. If instead they were released at a distance D/2, the acceleration of one charge would be ____.

A. 2a
B. 4a
C. a/2
D. a/4

Answer 1

B. 4a

Step-by-step explanation:

Force between the charges is inversely proportional to the square of the distance

=> Force will be 4 times and acceleration will be 4a

=> Answer b).

Answer 2

Answer: B. 4a

Explanation: The force experienced by the two identical charge is related to thier acceleration a by

F = m*a

m is mass of charge

a is thier acceleration

Also,

Force F is inversely proportional to D². From Coulomb's law

So we have

F ~1/D²

Since the distance D was reduced to D/2

We have that

F¹ ~ 1/(D/2)²

F¹~ 1/D²/4

From the law of reciprocal we have

F¹~ 4*1/D²

Which is the new force due to the decrease in distance to D/2

But 1/D² is original force F.

F¹ ~ 4*F

But F= m*a

F¹ ~ 4*m*a

But m is constant so the only thing changing is the acceleration. So we have, the new acceleration increased by a factor of 4 to be 4*a = 4a.