Question

If two charged particles (the charge on each is \(Q\) ) are separated by a distance \(d\), there is a force \(F\) between them. What is the force if the magnitude of each charge is doubled and the distance between them changes to \(2 d ?\)

Short answer

**Answer:** The force between the two charged particles when the magnitude of each charge is doubled and the distance between them changes to 2d is twice the initial force, i.e. \(F'= 2F\).

Step-by-step solution

Understand Coulomb's Law

Coulomb's law states that the force between two charged particles is given by: \(F =k\frac{q_1q_2}{r^2}\), where \(k\) is the Coulomb constant, \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between the charges. 2. Identify Initial Variables

Identify Initial Variables

In our initial setup, we have charges \(Q\) and a distance \(d\). Thus, we can rewrite the force formula as: \(F = k\frac{QQ}{d^2}\) 3. Identify Changed Variables

Identify Changed Variables

Now, we are asked to find the force when the charges are doubled and the distance between them changes to \(2d\). So now we have charges \(2Q\) and distance \(2d\). 4. Calculate New Force

Calculate New Force

We substitute the new charges and distance into Coulomb's law: \(F' = k\frac{(2Q)(2Q)}{(2d)^2}\) 5. Simplify the Expression

Simplify the Expression

Now, simplify the expression for the new force: \(F' = k\frac{4Q^2}{4d^2}\) 6. Compare New Force to Initial Force

Compare New Force to Initial Force

The initial force was given by \(F = k\frac{QQ}{d^2}\), and the new force is \(F' = k\frac{4Q^2}{4d^2}\). To find the relationship between the new force and the initial force, we can rewrite the new force as: \(F' = 2\times \left(k\frac{QQ}{d^2}\right)\) 7. State the Final Answer

State the Final Answer

The force between the two charged particles when the magnitude of each charge is doubled and the distance between them changes to \(2 d\) is twice the initial force, i.e. \(F'= 2F\).