Electric Potential And Potential Difference

Step 1: Relate the charge and radius of the small and big spheres.
Let be the number of small spheres, . Let be the radius of a small sphere and be the radius of the big sphere. The total volume remains constant when the spheres combine (assuming no loss). For , we have . So .

Step 2: Relate the charge of the small and big spheres.
The total charge is conserved. Let be the charge of a small sphere and be the charge of the big sphere.

Step 3: Relate the potential to charge and radius.
The electric potential of a charged conducting sphere is . Potential of the small sphere: . Potential of the big sphere: .

Step 4: Express the potential of the big sphere in terms of the small sphere's potential.


Step 5: Calculate the potential of the big sphere.
Substitute and . Since , we have .

Step 1: Analyze the initial state (with dielectric, connected to battery).
Capacitance with air, F.
Dielectric constant, .
Capacitance with dielectric, F.
The capacitor is charged to a potential V.
The charge stored on the capacitor is C.
The energy stored initially is J.
Step 2: Analyze the final state (dielectric removed, battery disconnected).
The battery is disconnected, so the charge on the plates remains constant. C.
The dielectric is removed, so the capacitance reverts to the capacitance with air, F.
The energy stored in this final state is .
J.
Step 3: Calculate the work done.
The work done by an external agent to remove the dielectric is equal to the change in the potential energy stored in the capacitor.
Work Done = .
Work Done = J.
Positive work is done because the electric field pulls the dielectric in, so an external agent must do work to pull it out against this force.

Step 1: Understanding the Concept:
This problem applies the Work-Energy Theorem. When a charged particle is accelerated by an electric field through a potential difference, the work done by the electric field on the particle is converted into its kinetic energy.
Step 2: Key Formula or Approach:
The work done ( ) by the electric field on a charge ( ) moving through a potential difference ( ) is given by:
The change in kinetic energy ( ) of the particle is:
According to the Work-Energy Theorem, .
Step 3: Detailed Explanation:
First, we list the given values in SI units:
- Mass, .
- Charge, .
- Potential difference, .
- The particle starts from rest, so its initial velocity, .
Now, we apply the Work-Energy Theorem:
We rearrange the formula to solve for the final velocity, :
Substitute the given values:
The terms in the numerator and denominator cancel out.
Now, take the square root to find the velocity:
Step 4: Final Answer:
The velocity acquired by the particle is 8 ms⁻¹. Therefore, option (D) is correct.