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

The new acceleration is B) 4a

Explanation:

According to Newton's second law, the acceleration of an object is proportional to the force exerted on it:

$a=\frac{F}{m}$

where

F is the net force on the object

m is its mass

a is the acceleration

This means that we can analyze the problem in terms of the force involved. The electric force between the charges at the beginning is given by

$F=k\frac{q_1 q_2}{D^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

D is the separation between the two charges

Later, the charges are separated by a new distance

D' = D/2

Therefore, the new force between them is

$F'=k\frac{q_1 q_2}{(D/2)^2}=4(k\frac{q_1 q_2}{D^2})=4F$

So, the force has increased by a factor of 4: and since the acceleration is proportional to the force, this means that the acceleration has also increased by a factor of 4, so the correct choice is

B) 4a

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