Question

A parallel plate capacitor of capacitance C₁ with a dielectric slab in between its plates is connected to a battery. It has a potential difference V₁ across its plates. When the dielectric slab is removed, keeping the capacitor connected to the battery, the new capacitance and potential difference are C₂ and V₂ respectively. Then,

• A. V₁ > V₂, C₁ > C₂
• B. V₁ < V₂, C₁ > C₂
• C. V₁ = V₂, C₁ > C₂
• D. V₁ = V₂, C₁ < C₂

Answer 1

The Correct Option is C

A parallel plate capacitor with capacitance C₁ and a dielectric slab is connected to a battery, providing a potential difference V₁.

When the dielectric slab is removed while keeping the battery connected, the capacitance changes to C₂, and the new potential difference is V₂.

Understanding the Changes 

The capacitance of a parallel plate capacitor with a dielectric is given by: \[ C_1 = \kappa C_0 \] where \( \kappa \) is the dielectric constant and \( C_0 \) is the capacitance without the dielectric.

When the dielectric is removed, the new capacitance becomes: \[ C_2 = C_0 \] Since \( \kappa > 1 \), it follows that: \[ C_1 > C_2 \]

Effect on Voltage

Since the capacitor remains connected to the battery, the voltage across the plates is maintained at the same level: \[ V_1 = V_2 \]

Conclusion

From the above analysis: C₁ > C₂ and V₁ = V₂.

Hence, the correct answer is: V₁ = V₂, C₁ > C₂.

Answer 2

In the case of a parallel plate capacitor, the capacitance \( C \) is related to the dielectric constant \( k \) by the formula: \[ C = k C_0 \] where \( C_0 \) is the capacitance of the capacitor without the dielectric. - When the dielectric slab is inserted between the plates, the capacitance increases by a factor of the dielectric constant \( k \), and the potential difference across the plates \( V_1 \) is given by: \[ V_1 = \frac{Q}{C_1} \] where \( C_1 \) is the capacitance with the dielectric and \( Q \) is the charge on the capacitor. - When the dielectric is removed, the capacitance decreases, but the battery is still connected. This means the charge \( Q \) remains constant, and the potential difference across the plates increases. This gives a new potential difference \( V_2 \) and capacitance \( C_2 \) such that: \[ C_2 < C_1 \quad \text{and} \quad V_2 > V_1 \] Thus, we have the relationship: \[ V_1 < V_2, \quad C_1 > C_2 \] \noindent

\(\textbf{Correct Answer:}\) (C) \( V_1 = V_2, C_1 > C_2 \)