Question

Two positive point charges Q and 2Q are separated by a distance R. If the charge Q experiences a force of magnitude F when the separation is R, what is the magnitude of the force on the charge 2Q when the separation is 2R

Answer 1

Answer:

$F'=\frac{1}{4}F$

Explanation:

According to Coulomb's law, the magnitude of each of the electric forces experiences by the two point charges is directly proportional to the product of the magnitude of both charges ($Q,2Q$) and inversely proportional to the square of the distance(R) that separates them:

$F=\frac{kQ(2Q)}{R^2}$

We have $R'=2R$

$F'=\frac{kQ(2Q)}{R'^2}\\F'=\frac{kQ(2Q)}{(2R)^2}\\F'=\frac{1}{4}\frac{kQ(2Q)}{R^2}\\F'=\frac{1}{4}F$

Answer 2

The magnitude of the force on the charge 2Q when the separation is 2R is F' = 1\4F

Coulomb's law:

As per this law, the magnitude for each & every of the electric force should be experience via the two-point charges also it should be directly proportional to the product with respect to the magnitude of the both charges. And, it should be an inverse proportional to the distance square.

So,

F = KQ(2Q) / R'

Here R' = 2R

So,

F' = KQ(2Q) / R'^2

= KQ(2Q) / (2R)^2

= 1/4 KQ(2Q)/R^2

= 1/4 F

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