Question

Two point charges, Q1 and Q2, are separated by a distance R. If the magnitudes of both charges are halved and their separation is also halved, what happens to the electrical force that each charge exerts on the other one?

a.) It increases by a factor of 8.
b.) It remains the same.
c.) It increases by a factor of 2.
d.) It increases by a factor of 16.
e.) It increases by a factor of 4.

Answer 1

Answer:

option B

Explanation:

given,

Two point charges, Q₁ and Q₂

distance between charge = R

Force between charges

$F = \dfrac{kQ_1Q_2}{R^2}$

now,

Separation between charges is half

The magnitudes of both charges are halved.

$F' = \dfrac{kQ'_1Q'_2}{R'^2}$

$F' = \dfrac{k\dfrac{Q_1}{2}\dfrac{Q_2}{2}}{(\dfrac{R}{2})^2}$

$F = \dfrac{\dfrac{1}{4}}{\dfrac{1}{4}}\dfrac{kQ_1Q_2}{R^2}$

$F' = F$

Hence, the force remain same.

The correct answer is option B