Problem 56 - Master Resource Book in JEE Main Physics

Question

Two point charges placed at a certain distance \(r\) in air exert a force \(F\) on each other. Then the distance \(r\) at which these charges will exert the same force in a medium of dielectric constant \(K\) is given by (a) \(r / K\) (b) \(r / \sqrt{K}\) (c) \(r \sqrt{K}\) (d) \(r\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the Original Force Equation

The force between two point charges in air is given by Coulomb's Law: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \] where \(k\) is Coulomb's constant, \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between them.

Step 2: Effect of Dielectric Medium

When the charges are placed in a medium with a dielectric constant \(K\), the force is given by: \[ F_{\text{medium}} = \frac{k \cdot |q_1 \cdot q_2|}{(r^\prime)^2 \cdot K} \] where \(r^\prime\) is the new distance in the medium, and we want this force to equal the original force \(F\).

Step 3: Equating Forces in Air and Medium

Set the two force equations equal to each other to find \(r^\prime\): \[ \frac{k \cdot |q_1 \cdot q_2|}{r^2} = \frac{k \cdot |q_1 \cdot q_2|}{(r^\prime)^2 \cdot K} \]

Step 4: Solving for the New Distance

Simplify and solve the equation for \(r^\prime\): \[ \frac{1}{r^2} = \frac{1}{K \cdot (r^\prime)^2} \] Cross-multiply to get: \[ (r^\prime)^2 = K \cdot r^2 \] Taking the square root gives us: \[ r^\prime = r \sqrt{K} \]

Step 5: Conclusion

Therefore, the distance \(r\) at which the charges will exert the same force in a medium with dielectric constant \(K\) is \( r \sqrt{K} \). This corresponds to option (c).

Key Concepts

Dielectric Constant

The dielectric constant, often denoted by the letter \( K \), plays a pivotal role in electromagnetism and is crucial to understanding how materials affect electric fields. It is a dimensionless quantity that measures a material's ability to reduce the electric field within it compared to the vacuum. In simpler terms, it tells us how much an electric field is "weakened" when it passes through a particular medium versus air or vacuum.

• A vacuum has a dielectric constant of 1.
• Most materials have a dielectric constant greater than 1, which means they can weaken the electric field more than air does.

Materials with higher dielectric constants are more effective at reducing electric fields, thereby affecting the force between charges when they are placed in such mediums.

Force Between Point Charges

Coulomb's Law is the foundational principle for understanding the force between two point charges. This law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, this can be represented as:
\[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\] where \( F \) is the force, \( q_1 \) and \( q_2 \) are the charges, \( r \) is the distance between the charges, and \( k \) is Coulomb's constant.

• The force is attractive if charges have opposite signs and repulsive if they are like charges.
• Two charges twice as far apart will experience a force four times weaker, due to the \( 1/r^2 \) relationship.
• The force magnitude is unchanged if the magnitude of charges or the square of the distance remains constant.

This law provides a direct way to calculate the magnitude of the electrostatic force between two point charges.

Distance in Dielectric Medium

Understanding how distance between charges changes in a dielectric medium involves rearranging Coulomb's Law with the dielectric constant in place. When point charges are placed in a medium with dielectric constant \( K \), the force between them is reduced. The modified equation becomes:
\[F_{\text{medium}} = \frac{k \cdot |q_1 \cdot q_2|}{K \cdot (r')^2}\]
Here, \( r' \) is the adjusted distance in the medium. To find this new distance where the force in a dielectric is equal to that in air, solve:
\[\frac{1}{r^2} = \frac{1}{K \cdot (r')^2}\]
By cross-multiplying and solving for \( r' \), we find:
\[(r')^2 = K \cdot r^2\] \[r' = r \sqrt{K}\]

• This equation explains how the distance changes for the same force to act in a dielectric medium like it does in air.
• The factor \( \sqrt{K} \) shows that distance has to be increased in a medium with dielectric constant \( K \) to maintain the same force level.

The revised distance depends on both the initial separation and the dielectric constant.