Understanding Electrostatic Force and Distance

The question asks how the electrostatic force between two charged particles changes when the distance between them is significantly reduced. This involves applying Coulomb's Law, which describes the force between point charges.

What is Coulomb's Law?

Coulomb's Law states that the electrostatic force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. Mathematically, it is given by:

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

Where:

• \(F\) is the magnitude of the electrostatic force.
• \(k\) is Coulomb's constant (a proportionality constant).
• \(q_1\) and \(q_2\) are the magnitudes of the two charges.
• \(r\) is the distance between the two charges.

The key aspect here is the inverse square relationship with distance (\(r^2\)). This means if the distance changes, the force changes by the inverse square of that change factor.

Analyzing the Problem Setup

Let's denote the initial situation with subscript 1 and the new situation with subscript 2.

• Initial force = \(F_1 = F\)
• Initial distance = \(r_1\)
• New distance = \(r_2\)
• New force = \(F_2\)

According to the question, the new distance is reduced to one-fifth of the original distance. So, \(r_2 = \frac{1}{5} r_1\).

The charges \(q_1\) and \(q_2\) remain the same.

Calculating the New Force

Using Coulomb's Law for both situations:

Initial Force: \(F_1 = k \frac{|q_1 q_2|}{r_1^2}\)

New Force: \(F_2 = k \frac{|q_1 q_2|}{r_2^2}\)

We can find the ratio of the new force to the initial force:

\(\frac{F_2}{F_1} = \frac{k \frac{|q_1 q_2|}{r_2^2}}{k \frac{|q_1 q_2|}{r_1^2}}\)

The constant \(k\) and the charge product \(|q_1 q_2|\) cancel out, leaving:

\(\frac{F_2}{F_1} = \frac{r_1^2}{r_2^2}\)

Now, substitute the relationship \(r_2 = \frac{1}{5} r_1\):

\(\frac{F_2}{F_1} = \frac{r_1^2}{(\frac{1}{5} r_1)^2} = \frac{r_1^2}{\frac{1}{25} r_1^2}\)

The \(r_1^2\) terms cancel out:

\(\frac{F_2}{F_1} = \frac{1}{\frac{1}{25}} = 25\)

So, \(F_2 = 25 \times F_1\).

Since the initial force \(F_1\) is given as \(F\), the new force \(F_2\) is \(25F\).

This result shows that when the distance between the charges is reduced to one-fifth, the electrostatic force increases by a factor of \(5^2 = 25\).

Conclusion

When the distance between the two charges is reduced to one-fifth, the electrostatic force between them becomes 25 times the original force.

Revision Table: Coulomb's Law and Force

Additional Information: Electrostatic Force

Electrostatic force is a fundamental force in nature, responsible for attractions and repulsions between electrically charged particles. It is an example of a conservative force. Coulomb's Law is analogous to Newton's Law of Universal Gravitation, both following an inverse square law with distance, though electrostatic force can be attractive or repulsive, while gravitational force is always attractive.

The constant \(k\) in vacuum is approximately \(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\). In a medium other than vacuum, the force is reduced by a factor called the dielectric constant of the medium.