Problem 36 - Master Resource Book in JEE Main Physics

Question

Two point charges exert on each other a force \(F\) when they are placed \(r\) distance apart in air. If they are placed \(R\) distance apart in a medium of dielectric constant \(K\), they exert the same force. The distance \(R\) equals (a) \(\frac{r}{K}\) (b) \(r \bar{K}\) (c) \(r \sqrt{K}\) (d) \(\frac{r}{\sqrt{K}}\)

Step-by-Step Solution

Answer

The correct answer is (d) \( \frac{r}{\sqrt{K}} \).

Step 1: Understand Coulomb's Law in Air

Coulomb's Law states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) in air (or vacuum) is given by the formula: \[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]where \( k \) is Coulomb's constant.

Step 2: Express Force in a Medium with Dielectric Constant

When the charges are placed in a medium with dielectric constant \( K \), the force \( F' \) between the two charges is given by: \[ F' = \frac{k \cdot q_1 \cdot q_2}{K \cdot R^2} \]where \( R \) is the new distance between them in the medium.

Step 3: Set Forces Equal and Simplify

Given that the forces are equal in both cases \( F = F' \), we have:\[ \frac{k \cdot q_1 \cdot q_2}{r^2} = \frac{k \cdot q_1 \cdot q_2}{K \cdot R^2} \]Simplifying, we divide through by \( k \cdot q_1 \cdot q_2 \):\[ \frac{1}{r^2} = \frac{1}{K \cdot R^2} \]

Step 4: Solve for R in Terms of r and K

Cross-multiplying the equation from step 3, we get:\[ K \cdot R^2 = r^2 \]Divide both sides by \( K \):\[ R^2 = \frac{r^2}{K} \]Taking the square root of both sides results in:\[ R = \frac{r}{\sqrt{K}} \]

Key Concepts

Dielectric Constant

A dielectric constant, often symbolized as \( K \), is a measure of a material's ability to resist the formation of an electric field within it. This constant represents how much the material can reduce the effective force between two point charges.
In simpler terms, it tells us how much "weaker" the electric field in a medium is compared to a vacuum or air.
For example:

• If \( K = 1 \), the medium doesn't alter the electric force.
• If \( K > 1 \), the medium reduces the force between charges.
• The higher the dielectric constant, the more the force between charges is reduced.

This property is crucial in understanding how different materials change the interactions between charged particles. Materials with higher dielectric constants are often used in electrical systems to isolate or insulate components, allowing for better control of electric interactions.

Force Between Charges

The force between two charges, according to Coulomb's Law, depends on several factors: the magnitude of each charge, the distance between them, and the medium the charges are in.
This law is usually expressed mathematically as:\[ F = \frac{k \cdot q_1 \cdot q_2}{r^2} \]where:

• \( F \) is the force between the charges.
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the distance separating the charges.
• \( k \) is Coulomb's constant, which has a fixed value in air.

The equation illustrates how larger charges generate a stronger force, and how increasing the distance between them reduces the force.
The force is inversely proportional to the square of the distance, meaning that even small changes in distance significantly affect the force. This foundational principle helps us understand magnetic and electric fields in various fields of physics.

Medium Effect on Electric Forces

The medium in which charges are placed significantly affects the electric force between them.
A medium with a dielectric constant greater than 1 reduces the electrostatic force acting between charges.
This is because the medium allows partial alignment of molecules in place, ultimately opposing and decreasing the net electric field.

• If charges are placed in a vacuum, the force is at its strongest.
• In materials with a high dielectric constant (like water), the force between the same charges is weaker than in air.

Understanding this concept is important when designing and using devices that rely on specific electric force conditions, like capacitors or insulators.
Recognizing how the medium modifies forces can help in predicting the behavior of electric fields and their interactions in practical use.

Electric Force Equation

The electric force equation changes when considering the impact of different media, captured by the dielectric constant.
In a medium with a dielectric constant \( K \), the equation for electric force transforms to: \[ F' = \frac{k \cdot q_1 \cdot q_2}{K \cdot R^2} \]This form shows us:

• The effective force \( F' \) is reduced by a factor proportional to \( K \).
• The distance \( R \) between charges is new and often different from when they are in air.

With this equation, you can calculate the force between any two charges placed in various mediums.
It reveals that both the nature of the medium and the distance impact how forces behave, and it helps bridge theoretical concepts with practical situations, where materials and distances are limited by real-world constraints.