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

Answer:

d) 4a

Explanation:

r = Distance

Electrostatic force is given by

$F=\dfrac{kq_1q_2}{r^2}$

$ma=\dfrac{kq_1q_2}{r^2}$

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

$a\propto \dfrac{1}{r^2}$

If the distance is reduced

$\dfrac{a_1}{a_2}=\dfrac{r_2^2}{r_1^2}\\\Rightarrow \dfrac{a_1}{a_2}=\dfrac{(\dfrac{d}{2})^2}{d^2}\\\Rightarrow \dfrac{a_1}{a_2}=\dfrac{1}{4}\\\Rightarrow a_2=4a_1$

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

The answer is d) 4a