Question

Two equal and opposite charges separated by a distance, d, experience an attractive force, F. If each charge is doubled in magnitude, what should be the new distance between them so that the force remains unchanged?

Answer 1

If the each charge is doubled in magnitude and force remains unchanged then the distance between the charged particle is four times the original distance.

Explanation:

The force between two charged objects are shown by Coulomb's law.

In the scalar form, this law is represented as:

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

where,

k = coulomb's constant

q₁ and q₂ are the magnitude of charged objects

d = distance between the charges.

According to the question,

Case1:

charge on object 1 = +q₁

Charge on object 2 = -q₂

Distance between the charges = d

$Force, F = k\frac{+q_1 X -q_2}{d^2} \\\\F = k\frac{-q_1q_2}{d^2}$

Case 2:

Charge on object 1 = +2q₁

Charge on object 2 = -2q₂

Force between the objects is same

New distance, dₓ = ?

If both the forces are equal then,

$k\frac{-q_1q_2}{d^2} = k\frac{+2q_1 X -2q_2}{d_x^2}$

On simplifying the equation we get,

$\frac{-1}{d^2} = \frac{-4}{d_x^2} \\\\d_x^2 = 4d^2$

So, if the each charge is doubled in magnitude and force remains unchanged then the distance between the charged particle is four times the original distance.