Effect of Dielectric on Capacitance - UNSOLVED PRACTICE SET
Chapter: Electrostatic Potential and Capacitance | Topic: Effect of Dielectric on Capacitance
EFFECT OF DIELECTRIC ON CAPACITANCE - UNSOLVED PRACTICE SET
Topic: Effect of Dielectric on Capacitance
Multiple Choice Questions
Q1. When a dielectric slab is inserted between the plates of a parallel plate capacitor while the battery remains connected:
- The capacitance decreases
- The capacitance increases by a factor K
- The charge remains constant
- The potential difference increases
Q2. When a dielectric slab is inserted between the plates of a charged capacitor after disconnecting the battery:
- The charge increases
- The potential difference decreases by a factor K
- The capacitance remains unchanged
- The energy increases
Q3. The dielectric constant K of a material is always:
- Less than 1
- Equal to 1
- Greater than or equal to 1
- Negative
Q4. For a parallel plate capacitor with a dielectric slab of thickness t (t < d) inserted between the plates, the capacitance:
- Decreases compared to air
- Increases compared to air
- Remains the same as air
- Becomes zero
Q5. When a dielectric is inserted in a capacitor connected to a battery, the energy stored:
- Decreases
- Increases by a factor K
- Remains the same
- Becomes zero
Q6. The dielectric constant of vacuum is:
- Zero
- 1
- Infinity
- Cannot be defined
Short Answer Questions
Q7. Explain why the capacitance of a parallel plate capacitor increases when a dielectric is inserted between its plates.
Q8. A parallel plate capacitor is charged by a battery and then the battery is removed. A dielectric slab is now inserted. What happens to (a) the charge, (b) the potential difference, and (c) the energy stored?
Q9. A parallel plate capacitor remains connected to a battery. A dielectric slab is inserted. What happens to (a) the charge, (b) the potential difference, and (c) the energy stored?
Q10. Why does the energy stored in a capacitor decrease when a dielectric is inserted after disconnecting the battery? Where does the "lost" energy go?
Q11. A parallel plate capacitor has air between its plates. Its capacitance is 10 pF. What will be its capacitance if a mica sheet of dielectric constant K = 6 completely fills the space between the plates?
Q12. A dielectric slab of thickness t and dielectric constant K is inserted between the plates of a parallel plate capacitor of plate separation d (t < d). Write the expression for the new capacitance.
Long Answer Questions
Q13. Derive the expression for the capacitance of a parallel plate capacitor when a dielectric slab of thickness t and dielectric constant K is inserted between the plates (t < d). Explain what happens when t = d.
Q14. Explain the effect of inserting a dielectric in a capacitor (a) when the battery remains connected, and (b) when the battery is disconnected before inserting the dielectric. Compare the changes in Q, V, C, and U in both cases using a table.
Q15. A parallel plate capacitor has plate area A = 100 cm² and plate separation d = 2 mm. It is connected to a 50 V battery.
(a) Calculate the initial capacitance, charge, and energy.
(b) A dielectric slab of K = 4 and thickness 2 mm is inserted while the battery remains connected. Calculate the new values.
(c) Now the battery is disconnected and the dielectric is removed. Calculate the new potential difference and energy.
[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²]
Numerical / Application-Based Problems
Q16. A parallel plate capacitor has plates of area 400 cm² separated by 3 mm in air. It is charged to 120 V and then isolated from the battery.
(a) Calculate the initial capacitance, charge, and energy.
(b) A dielectric slab of K = 5 and thickness 2 mm is inserted between the plates. Calculate the new capacitance, potential difference, and energy.
(c) If instead, the dielectric completely fills the space (thickness = 3 mm), calculate the new values.
(d) Compare the energy in parts (b) and (c) and explain which case stores more energy and why.
[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²]
Q17. In your mobile phone, the touchscreen works using capacitive sensing. The screen can be modeled as a parallel plate capacitor with area A = 50 cm², initial separation d = 1 mm, and dielectric constant K = 3 (glass screen).
(a) Calculate the capacitance of the screen.
(b) When you touch the screen with your finger, the effective separation decreases to 0.8 mm. Calculate the change in capacitance.
(c) If the screen is charged to 5 V, calculate the change in charge and energy when you touch it.
(d) The phone detects this change in capacitance to register your touch. Explain why this technology is called "capacitive touch sensing."
[Given: ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻²]
Q18. A parallel plate capacitor has plate area A and plate separation d. It is connected to a battery of emf V. A dielectric slab of dielectric constant K and thickness t = d/2 is inserted between the plates.
(a) Derive the expression for the capacitance with the partially filled dielectric.
(b) Calculate the ratio of the new capacitance to the original capacitance.
(c) Find the electric field in the air gap and in the dielectric.
(d) Calculate the bound charge density on the dielectric surfaces.
(e) If two such dielectric slabs, each of thickness d/2 and dielectric constants K₁ and K₂, are placed in series between the plates, derive the equivalent capacitance.