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Two charges equal in magnitude and opposite in polarity are placed at a certain distance apart and force acting between them is F. If 75% charge of one is transferred to another, then the force between charges becomes

Prabhmeet , 8 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

To understand how the force between two charges changes when one charge is altered, we can start by recalling Coulomb's law, which describes the force between two point charges. The law states that the force \( F \) between two charges \( q_1 \) and \( q_2 \) separated by a distance \( r \) is given by the formula:

Coulomb's Law

The formula can be expressed as:

F = k * |q1 * q2| / r²

Here, \( k \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between them.

Initial Setup

Let's denote the initial charges as \( q \) and \( -q \). The force acting between these two charges is:

F = k * |q * (-q)| / r² = k * q² / r²

Charge Transfer Scenario

Now, if we transfer 75% of the charge from one charge to the other, we need to calculate the new charges. When 75% of charge \( q \) is transferred from \( q \) to \( -q \), the new charges become:

  • New charge on the first charge: \( q' = q - 0.75q = 0.25q \)
  • New charge on the second charge: \( q'' = -q + 0.75q = -0.25q \)

Calculating the New Force

Now, we can find the new force \( F' \) between these modified charges:

F' = k * |q' * q''| / r²

Substituting the new charges into the equation gives:

F' = k * |(0.25q) * (-0.25q)| / r² = k * (0.25q)² / r²

This simplifies to:

F' = k * (0.0625q²) / r²

Comparing Forces

Now, let's compare the new force \( F' \) with the original force \( F \):

F' = 0.0625 * (k * q² / r²) = 0.0625 * F

This means that the new force is 6.25% of the original force. Therefore, when 75% of the charge from one of the charges is transferred to the other, the force between the charges decreases significantly.

Summary

In summary, transferring 75% of one charge to another results in a new force that is only 6.25% of the original force. This example illustrates how sensitive the electrostatic force is to changes in charge magnitude, highlighting the inverse relationship between charge and force in electrostatics.

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