NEET Electrostatic Potential and Capacitance — Set 22

Question 1

A dielectric slab is partially inserted between the plates of a parallel plate capacitor while maintaining a constant voltage across the plates. What happens to the electric field between the plates in the region where the dielectric is present?

Accepted Answer

When a dielectric slab (with dielectric constant $ K > 1 $) is inserted between the plates of a capacitor with constant voltage $ V $, the electric field $ E $ in the dielectric region decreases. The electric field in a dielectric is given by $ E = \frac{E_0}{K} $, where $ E_0 = \frac{V}{d} $ is the field without the dielectric ($ d $ is the plate separation). Since $ K > 1 $, $ E $ is reduced in the dielectric region compared to the air region, as the dielectric polarizes and induces charges that oppose the applied field.

Question 2

In a system of two identical conductors initially charged differently and then connected by a wire, why does the final energy of the system decrease?

Accepted Answer

Initially, the conductors have charges $ Q_1 $ and $ Q_2 $, with energy $ U_i = \frac{Q_1^2}{2C} + \frac{Q_2^2}{2C} $. Upon connection, charge redistributes to equal potentials, total charge $ Q_1 + Q_2 $ splits equally ($ \frac{Q_1 + Q_2}{2} $ each), so final energy $ U_f = 2 \times \frac{\left(\frac{Q_1 + Q_2}{2}\right)^2}{2C} = \frac{(Q_1 + Q_2)^2}{4C} $. Typically, $ (Q_1 + Q_2)^2 < Q_1^2 + Q_2^2 $ unless $ Q_1 = Q_2 $, so $ U_f < U_i $, with energy lost as heat or radiation during charge flow.

Question 3

A point charge $ Q = 18 \times 10^{-9} \, \text{C} $ is placed at the origin. Calculate the potential at a point 6 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{18 \times 10^{-9}}{6} = 9 \times 10^9 \times 3 \times 10^{-9} = 27 \, \text{V} $.

NEET Physics: Electrostatic Potential and Capacitance — Practice Set 22

Q1. A dielectric slab is partially inserted between the plates of a parallel plate capacitor while maintaining a constant voltage across the plates. What happens to the electric field between the plates in the region where the dielectric is present?

Q2. In a system of two identical conductors initially charged differently and then connected by a wire, why does the final energy of the system decrease?

Q3. A point charge \( Q = 18 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q4. Five capacitors of \( 20 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?

Q5. A dipole \( p = 2 \times 10^{-9} \, \text{C m} \) is rotated from \( \theta = 90^\circ \) to \( 0^\circ \) in a field \( E = 5 \times 10^5 \, \text{N/C} \). What is the work done?

Q7. Three capacitors \( 5 \, \text{pF} \), \( 10 \, \text{pF} \), and \( 20 \, \text{pF} \) are in parallel. What is the total capacitance?

Q8. A point charge \( Q = 9 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 3 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q9. A parallel plate capacitor with \( C = 70 \, \text{pF} \) in air has a dielectric (\( K = 5 \)) inserted fully between plates. What is the new capacitance?

Q10. A dipole with \( p = 4 \times 10^{-9} \, \text{C m} \) makes an angle of \( 30^\circ \) with a uniform field \( E = 5 \times 10^4 \, \text{N/C} \). What is its potential energy?