Question

Just after two identical point charges are released when they are a distance D apart in outer space, they have an acceleration a. If you release them from a distance D/2 instead, their acceleration will be?

a) a/4 b) a/2 c) 2a d) 4a

Answer 1

d) 4a

Step-by-step explanation:

r = Distance

Electrostatic force is given by

`F=(kq_1q_2)/(r^2)`

`ma=(kq_1q_2)/(r^2)`

It can be seen that the force is inversely proportional to distance

`a\propto (1)/(r^2)`

If the distance is reduced

`(a_1)/(a_2)=(r_2^2)/(r_1^2)\\\Rightarrow (a_1)/(a_2)=(((d)/(2))^2)/(d^2)\\\Rightarrow (a_1)/(a_2)=(1)/(4)\\\Rightarrow a_2=4a_1`

So, the new acceleration will be four times the old acceleration

The answer is d) 4a