Question

When a capacitor of capacity \( C \) is charged with charge \( q \) and potential difference \( V \), the electric field \( E \) and the electrostatic energy \( U \) is obtained between plates of capacitor. If the capacitor is disconnected from the source and a slab of medium having the same thickness and dielectric constant \( k \) is introduced completely between the plates, calculate the new values of the following:
(i)Capacity
(ii) Potential difference
(iii) Charge
(iv) Electric field
(v) Electrostatic potential energy
(i) Capacity

Hint:

When a dielectric is inserted, the capacitance increases by a factor of \( k \), while the potential and electric field decrease proportionally.

Answer 1

Step 1: The capacitance of a parallel plate capacitor with dielectric is given by: \[ C' = kC \] \[ \boxed{C' = kC} \] (ii) Potential Difference % Correct Answer Correct Answer: \( V' = \frac{V}{k} \) % Solution Solution: Step 1: Since the capacitor is disconnected, the charge remains constant. Using the formula: \[ V' = \frac{q}{C'} = \frac{q}{kC} \] \[ V' = \frac{V}{k} \] \[ \boxed{V' = \frac{V}{k}} \] (iii) Charge % Correct Answer Correct Answer: \( q' = q \) (remains constant) % Solution Solution: Step 1: As the capacitor is disconnected, the charge remains the same: \[ q' = q \] \[ \boxed{q' = q} \] (iv) Electric Field % Correct Answer Correct Answer: \( E' = \frac{E}{k} \) % Solution Solution: Step 1: The electric field is given by: \[ E = \frac{V}{d} \] Step 2: Substituting the new potential difference: \[ E' = \frac{V'}{d} = \frac{V}{kd} \] \[ E' = \frac{E}{k} \] \[ \boxed{E' = \frac{E}{k}} \] (v) Electrostatic Potential Energy % Correct Answer Correct Answer: \( U' = \frac{U}{k} \) % Solution Solution: Step 1: Electrostatic potential energy is given by: \[ U = \frac{1}{2} C V^2 \] Step 2: Substituting new values: \[ U' = \frac{1}{2} kC \left( \frac{V}{k} \right)^2 \] \[ U' = \frac{U}{k} \] \[ \boxed{U' = \frac{U}{k}} \]