Two point charges in air are 2 m apart.One charge is three times the other charge.

1. Break Down the Problem

To analyze the problem involving two point charges, we identify the following key components:

1. Charge $q_1$ (let’s denote it as $q$).
2. Charge $q_2$ which is three times charge $q_1$, so $q_2 = 3q$.
3. The distance between the two charges $r = 2$ m.

2. Relevant Concepts

We will utilize Coulomb’s Law, which states that the electric force $F$ between two point charges is given by the formula:

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

where:

• $F$ is the magnitude of the force between the charges,
• $k$ is Coulomb's constant ($k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2$),
• $q_1$ and $q_2$ are the amounts of the charges,
• $r$ is the distance between the centers of the two charges.

3. Analysis and Detail

To find the force between the two charges, we substitute $q_1 = q$ and $q_2 = 3q$ into Coulomb’s Law:

$F = k \frac{|q \cdot (3q)|}{(2)^2}$ This simplifies to:

$F = k \frac{3q^2}{4}$

4. Verify and Summarize

In summary, the force $F$ between the two point charges, in terms of charge $q$, is given by:

$F = \frac{3kq^2}{4}$

Final Answer

The force between the two point charges can be expressed as:

$F = \frac{3 \cdot 8.99 \times 10^9 \cdot q^2}{4} \quad \text{Newton}$

To find a numerical value for the force, the charge $q$ must be specified.