All Exams Test series for 1 year @ ₹349 only
Question

Two point charges placed a distance d apart in vacuum exert a force of magnitude F on each other. One of the two charges is doubled. To keep the magnitude of force same the separation between the charges should be changed to

The correct answer is
$\sqrt{2}$ d

Understanding Electrostatic Force Calculation

This problem involves calculating the electrostatic force between two point charges. We need to determine how the distance between the charges must change to keep the force constant when one of the charges is modified. The fundamental principle governing this interaction is Coulomb's Law.

Coulomb's Law Explained

Coulomb's Law states that the magnitude of the electrostatic force ($F$) between two point charges ($q_1$ and $q_2$) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance ($r$) between them. In a vacuum, the formula is:

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

Where $k$ is Coulomb's constant ($k \approx 8.9875 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2$).

Initial Scenario Setup

Let the initial charges be $q_1$ and $q_2$, and the initial separation distance be $d$. The initial force between them is given as $F$. Using Coulomb's Law:

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

Modified Scenario and Force Constraint

The problem states that one of the charges is doubled. Let's assume $q_1$ is doubled, so the new charge becomes $2q_1$. The other charge, $q_2$, remains the same. We need to find the new separation distance, let's call it $d'$, such that the magnitude of the force remains the same, i.e., $F' = F$.

The new force $F'$ is given by:

$F' = k \frac{|(2q_1) q_2|}{(d')^2}$

Calculating the New Separation Distance

We are given that the magnitude of the force must remain the same ($F' = F$). Therefore, we can set the expressions for the forces equal to each other:

$k \frac{|2q_1 q_2|}{(d')^2} = k \frac{|q_1 q_2|}{d^2}$

Now, we can simplify the equation by canceling out common terms ($k$, $|q_1 q_2|$):

$\frac{2}{(d')^2} = \frac{1}{d^2}$

To solve for $d'$, we can rearrange the equation:

$(d')^2 = 2 d^2$

Taking the square root of both sides to find the new distance $d'$:

$d' = \sqrt{2 d^2}$

$d' = \sqrt{2} d$

This shows that to keep the force magnitude the same when one charge is doubled, the separation distance must be increased by a factor of $\sqrt{2}$.

Was this answer helpful?

Important Questions from Electric Charges and Fields

  1. Which of the following options is correct by using Coulomb's law?

  2. Which of the following statements are correct?

    • A. The angle at minimum deviation of a prism is greater for violet light than that for red light.
    • B. The purpose of microscopes and telescopes is to increase the visual angle.
    • C. For the diffraction to take place, the size of aperture or of the obstacle should be comparable to the wavelength of light.
    • D. The light scattered in the direction of the incident light is always plane polarized.
    • E. The source and its virtual image can behave as coherent sources.

    Choose the correct answer from the options given below:

  3. Match List - I with List - II

    Choose the correct answer from the options given below:

  4. A thin metallic spherical shell contains a charge +10 μC on it. A point charge +2 μC is placed at the centre of the shell and another charge +5 μC is placed outside it as shown. The force on the charge +2 μC at the centre is:

  5. In the figure, an α-particle moves a distance l in a uniform electric field E as shown. Does the Electric Field do a positive or a negative work on the α-particle? Does the electric potential energy of the α-particle increase or decrease?

Need Expert Advice?

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App