NEET Electrostatic Potential and Capacitance — Set 15

Question 1

Why can't the electric field inside a conductor in electrostatic equilibrium have a tangential component along its surface?

Accepted Answer

In electrostatic equilibrium, the electric field inside a conductor is zero, and charges reside on the surface. If the electric field had a tangential component along the surface, it would exert a force on the free charges, causing them to move along the surface. This movement would contradict the condition of equilibrium where no net motion of charges occurs. Therefore, the field must be normal to the surface, ensuring no tangential component exists to drive charge motion.

Question 2

A point charge $ Q = 12 \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{12 \times 10^{-9}}{6} = 9 \times 10^9 \times 2 \times 10^{-9} = 18 \, \text{V} $.

Question 3

A capacitor with $ C = 10 \, \text{pF} $ in air has a dielectric ($ K = 4 $, thickness $ d/2 $) inserted. What is the new capacitance if $ d $ is the original separation? (Take $ \varepsilon_0 A/d = 10 \, \text{pF} $).

Accepted Answer

Potential difference: $ V = E_0 \left( \frac{d}{2} \right) + \frac{E_0}{K} \left( \frac{d}{2} \right) = E_0 d \left( \frac{1}{2} + \frac{1}{2 \times 4} \right) = E_0 d \left( \frac{1}{2} + \frac{1}{8} \right) = \frac{5}{8} E_0 d $. $ C = \frac{Q}{V} = \frac{Q}{\frac{5}{8} V_0} = \frac{8}{5} \times \frac{Q}{V_0} = \frac{8}{5} \times 10 = 16 \, \text{pF} $.

NEET Physics: Electrostatic Potential and Capacitance — Practice Set 15

Q1. Why can't the electric field inside a conductor in electrostatic equilibrium have a tangential component along its surface?

Q2. A point charge \( Q = 12 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 6 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q3. A capacitor with \( C = 10 \, \text{pF} \) in air has a dielectric (\( K = 4 \), thickness \( d/2 \)) inserted. What is the new capacitance if \( d \) is the original separation? (Take \( \varepsilon_0 A/d = 10 \, \text{pF} \)).

Q4. A \( 10 \, \mu\text{F} \) capacitor charged to \( 40 \, \text{V} \) is connected to an uncharged \( 10 \, \mu\text{F} \) capacitor. What is the final potential difference?

Q5. A conductor has a surface charge density of \( 4 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

Q6. A spherical conductor of radius 15 cm has a charge of \( 9 \times 10^{-8} \, \text{C} \). What is the electric field at 20 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q7. A parallel plate capacitor with capacitance \( 80 \, \text{pF} \) has a dielectric (\( K = 5 \), thickness \( d/5 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Q8. A dipole with \( p = 8 \times 10^{-9} \, \text{C m} \) makes an angle of \( 60^\circ \) with a uniform field \( E = 1 \times 10^5 \, \text{N/C} \). What is its potential energy?

Q9. A charge of \( 10 \, \mu\text{C} \) is moved from infinity to a point where the potential is \( 30 \, \text{V} \). What is the work done?

Q10. Four capacitors of \( 15 \, \mu\text{F} \) each are in series. What is the equivalent capacitance?