NEET Electrostatic Potential and Capacitance — Set 4

Question 1

Five capacitors of $ 25 \, \mu\text{F} $ each are in series. What is the equivalent capacitance?

Accepted Answer

$ \frac{1}{C} = \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25} + \frac{1}{25} = \frac{5}{25} $. $ C = \frac{25}{5} = 5 \, \mu\text{F} $.

Question 2

Why does the electric field between the plates of a parallel plate capacitor remain unchanged when the area of overlap between the plates is reduced while keeping the charge constant?

Accepted Answer

The electric field between the plates of a parallel plate capacitor is $ E = \frac{\sigma}{\varepsilon_0} $, where $ \sigma = \frac{Q}{A} $ is the surface charge density, $ Q $ is the charge, and $ A $ is the overlapping area. If the area $ A $ is reduced while $ Q $ is constant, $ \sigma $ increases ($ \sigma = \frac{Q}{A} $), but in the region of overlap, $ E = \frac{\sigma}{\varepsilon_0} $ depends on $ \sigma $, and adjustments occur such that $ E $ remains uniform between the overlapping plates. However, the field remains $ E = \frac{V}{d} $ in practice if plates are adjusted, but assuming ideal infinite plates approximation, $ E $ is determined solely by $ \sigma $, and in this context, charge redistribution keeps $ E $ unchanged locally.

Question 3

A dipole $ p = 6 \times 10^{-9} \, \text{C m} $ is rotated from $ \theta = 0^\circ $ to $ 90^\circ $ in a field $ E = 3 \times 10^5 \, \text{N/C} $. What is the work done?

Accepted Answer

Work done: $ W = p E (\cos \theta_0 - \cos \theta_1) = 6 \times 10^{-9} \times 3 \times 10^5 \times (\cos 0^\circ - \cos 90^\circ) $. $ W = 6 \times 10^{-9} \times 3 \times 10^5 \times (1 - 0) = 1.8 \times 10^{-3} \, \text{J} $.

NEET Physics: Electrostatic Potential and Capacitance — Practice Set 4

Q1. Five capacitors of \( 25 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

Q2. Why does the electric field between the plates of a parallel plate capacitor remain unchanged when the area of overlap between the plates is reduced while keeping the charge constant?

Q3. A dipole \( p = 6 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 0^\circ \) to \( 90^\circ \) in a field \( E = 3 \times 10^5 \, \text{N/C} \). What is the work done?

Q4. A parallel plate capacitor has plates of area \( 0.1 \, \text{m}^2 \) and separation 0.5 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

Q5. A \( 3 \, \mu\text{F} \) capacitor charged to \( 150 \, \text{V} \) is connected to an uncharged \( 3 \, \mu\text{F} \) capacitor. What is the final potential difference?

Q6. Two capacitors of \( 15 \, \text{pF} \) and \( 30 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

Q7. In a system of two identical capacitors in series, why does removing one capacitor while maintaining the same total voltage across the system increase the energy stored?

Q8. Two charges \( 12 \, \mu\text{C} \) and \( -6 \, \mu\text{C} \) are placed 15 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q9. A parallel plate capacitor has plates of area \( 0.04 \, \text{m}^2 \) and separation 0.2 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

Q10. A \( 4 \, \mu\text{F} \) capacitor charged to \( 300 \, \text{V} \) is connected to an uncharged \( 8 \, \mu\text{F} \) capacitor. What is the energy lost?