NEET Electrostatic Potential and Capacitance — Set 8

Question 1

Two charges $ 8 \, \mu\text{C} $ and $ -4 \, \mu\text{C} $ are at $ (6, 0, 0) $ and $ (-6, 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} $).

Accepted Answer

Distance to midpoint = 0.06 m. $ V = 9 \times 10^9 \left( \frac{8 \times 10^{-6}}{0.06} + \frac{-4 \times 10^{-6}}{0.06} \right) = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.06} $. $ V = 9 \times 10^9 \times \frac{4 \times 10^{-6}}{0.06} = 6 \times 10^5 \, \text{V} $.

Question 2

Two charges $ 10 \, \mu\text{C} $ and $ -5 \, \mu\text{C} $ are placed 20 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{10 \times 10^{-6} \times (-5 \times 10^{-6})}{0.2} = 9 \times 10^9 \times \frac{-50 \times 10^{-12}}{0.2} = -2.25 \, \text{J} $.

Question 3

Why does the capacitance of a parallel plate capacitor increase when a dielectric slab is inserted between the plates while keeping the plates connected to a battery?

Accepted Answer

When a dielectric slab (with $ K > 1 $) is inserted while the capacitor remains connected to a battery (constant voltage $ V $), the electric field decreases due to polarization ($ E = E_0/K $), and the capacitance increases ($ C' = K C $). This happens because the dielectric reduces the field, allowing more charge to be stored on the plates ($ Q = C V $, so $ Q' = K C V $) for the same voltage, effectively increasing the capacitance as $ C = \frac{Q}{V} $.

NEET Physics: Electrostatic Potential and Capacitance — Practice Set 8

Q1. Two charges \( 8 \, \mu\text{C} \) and \( -4 \, \mu\text{C} \) are at \( (6, 0, 0) \) and \( (-6, 0, 0) \, \text{cm} \). What is the potential at midpoint? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q2. Two charges \( 10 \, \mu\text{C} \) and \( -5 \, \mu\text{C} \) are placed 20 cm apart. What is the potential energy of the system? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q3. Why does the capacitance of a parallel plate capacitor increase when a dielectric slab is inserted between the plates while keeping the plates connected to a battery?

Q4. A metal sphere is placed inside a uniform electric field \( E \). What is the electric field inside the sphere after it reaches electrostatic equilibrium?

Q5. A parallel plate capacitor has plates of area \( 0.02 \, \text{m}^2 \) separated by 2 mm in air. What is its capacitance? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

Q6. A spherical conductor of radius 30 cm has a charge of \( 12 \times 10^{-8} \, \text{C} \). What is the electric field at 70 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q7. A point charge \( Q = 8 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 4 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q8. A parallel plate capacitor with \( C = 100 \, \text{pF} \) in air has a dielectric (\( K = 10 \)) inserted fully between plates. What is the new capacitance?

Q10. Why does the potential energy of a system of two opposite charges increase when their separation decreases?