NEET Electrostatic Potential and Capacitance — Set 25

Question 1

Two charges $ 32 \, \mu\text{C} $ and $ -16 \, \mu\text{C} $ are placed 32 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} $).

Accepted Answer

$ U = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r} = 9 \times 10^9 \times \frac{32 \times 10^{-6} \times (-16 \times 10^{-6})}{0.32} $. $ U = 9 \times 10^9 \times \frac{-512 \times 10^{-12}}{0.32} = -14.4 \, \text{J} $.

Question 2

Why does the energy density in an electric field depend on the square of the field strength?

Accepted Answer

The energy density in an electric field is given by $ u = \frac{1}{2} \varepsilon_0 E^2 $. This arises from the energy stored in a capacitor ($ U = \frac{1}{2} C V^2 $), scaled over the volume. For a parallel plate capacitor, $ E = \frac{V}{d} $, $ C = \frac{\varepsilon_0 A}{d} $, so $ U = \frac{1}{2} \frac{\varepsilon_0 A}{d} (E d)^2 = \frac{1}{2} \varepsilon_0 E^2 (A d) $, where $ A d $ is the volume. Thus, energy density $ u = \frac{U}{A d} = \frac{1}{2} \varepsilon_0 E^2 $, showing the $ E^2 $ dependence due to the quadratic relation between energy and field.

Question 3

A point charge $ Q = 5 \times 10^{-9} \, \text{C} $ is placed at the origin. What is the potential at a point 10 m away from the charge? (Take $ \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} $).

Accepted Answer

Potential due to a point charge: $ V = \frac{1}{4 \pi \varepsilon_0} \frac{Q}{r} $. Substitute: $ V = 9 \times 10^9 \times \frac{5 \times 10^{-9}}{10} = 9 \times 10^9 \times 0.5 \times 10^{-9} = 4.5 \, \text{V} $.

NEET Physics: Electrostatic Potential and Capacitance — Practice Set 25

Q1. Two charges \( 32 \, \mu\text{C} \) and \( -16 \, \mu\text{C} \) are placed 32 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q2. Why does the energy density in an electric field depend on the square of the field strength?

Q3. A point charge \( Q = 5 \times 10^{-9} \, \text{C} \) is placed at the origin. What is the potential at a point 10 m away from the charge? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q4. A parallel plate capacitor with \( C = 90 \, \text{pF} \) in air has a dielectric (\( K = 9 \)) inserted fully between plates. What is the new capacitance?

Q5. Why does the electric field near the edge of a charged conducting plate differ from the field at the center of the plate?

Q7. A \( 2 \, \mu\text{F} \) capacitor charged to \( 400 \, \text{V} \) is connected to an uncharged \( 6 \, \mu\text{F} \) capacitor. What is the energy lost?

Q8. A charge of \( 7 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 100 \, \text{V} \). What is the work done?

Q9. Two charges \( 16 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (4, 0, 0) \) and \( (-4, 0, 0) \, \text{cm} \). What is the potential at midpoint? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q10. An electric dipole with moment \( p = 8 \times 10^{-9} \, \text{C m} \) lies along the y-axis. What is the potential at \( (0, -4, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).