Question icon
Electrostatics

in a sphere with radius r from a distace of x a small particle of mass m and charge -q is placed at point p and released.an another charge Q attacs in it. if r>>x the particle will undergo oscillations along the axis of symmetry with an anguler frequency that is equal to?

Profile image of sagar singh
13 Years agoGrade
Answers icon

1 Answer

Profile image of Askiitians Tutor Team
ApprovedApproved Tutor Answer1 Year ago

To determine the angular frequency of oscillations for a small particle of mass \( m \) and charge \( -q \) placed at point \( P \) inside a sphere of radius \( r \), when another charge \( Q \) is introduced, we can use concepts from electrostatics and harmonic motion. Given that \( r \gg x \), we can simplify our analysis by assuming that the electric field created by charge \( Q \) is uniform across the region where the particle oscillates.

Understanding the Forces at Play

When the particle is released, it experiences a force due to the electric field generated by charge \( Q \). The force \( F \) acting on the particle can be expressed using Coulomb's law:

F = k \cdot \frac{|qQ|}{x^2}

Here, \( k \) is Coulomb's constant, and \( x \) is the distance from charge \( Q \) to the particle. Since the particle has a negative charge, it will be attracted towards the positive charge \( Q \).

Establishing the Potential Energy

The potential energy \( U \) associated with the particle in the electric field is given by:

U = -\frac{k \cdot qQ}{x}

As the particle moves closer to charge \( Q \), the potential energy decreases, indicating that the particle is moving towards a more stable configuration.

Finding the Effective Force and Oscillation

When the particle is displaced slightly from its equilibrium position, it will experience a restoring force that tends to bring it back to that position. This restoring force can be approximated as:

F_{\text{restoring}} = -k_{\text{eff}} \cdot x

Where \( k_{\text{eff}} \) is the effective spring constant. For small displacements, we can relate the force to the angular frequency \( \omega \) of oscillation:

F = m \cdot a = m \cdot \frac{d^2x}{dt^2}

By equating the restoring force to the mass times acceleration, we have:

m \cdot \frac{d^2x}{dt^2} = -k_{\text{eff}} \cdot x

Relating Angular Frequency to Spring Constant

The angular frequency \( \omega \) of the oscillation can be expressed as:

\(\omega = \sqrt{\frac{k_{\text{eff}}}{m}}\)

To find \( k_{\text{eff}} \), we can differentiate the potential energy with respect to \( x \) and evaluate it at the equilibrium position:

k_{\text{eff}} = -\frac{d^2U}{dx^2}

After performing the differentiation and substituting back into the equation for \( \omega \), we find that:

\(\omega = \sqrt{\frac{k \cdot |Q|}{m \cdot r^3}}\)

Final Expression for Angular Frequency

Thus, the angular frequency of the oscillations of the particle in the electric field created by charge \( Q \) is given by:

\(\omega = \sqrt{\frac{k \cdot |Q|}{m \cdot r^3}}\)

This formula indicates that the angular frequency depends on the charge \( Q \), the mass \( m \) of the particle, and the radius \( r \) of the sphere. As you can see, the larger the radius, the smaller the angular frequency, which aligns with our intuition that a larger radius leads to a weaker restoring force.