NEET Electrostatic Potential and Capacitance — Set 9

Question 1

A parallel plate capacitor with capacitance $ 60 \, \text{pF} $ has a dielectric ($ K = 4 $, thickness $ d/4 $) inserted. What is the new capacitance? (Original separation $ d $).

Accepted Answer

Potential difference: $ V = E_0 \left( \frac{3d}{4} \right) + \frac{E_0}{K} \left( \frac{d}{4} \right) = E_0 d \left( \frac{3}{4} + \frac{1}{4 \times 4} \right) = E_0 d \left( \frac{3}{4} + \frac{1}{16} \right) = E_0 d \times \frac{13}{16} $. $ C = \frac{Q}{V} = \frac{Q}{\frac{13}{16} V_0} = \frac{16}{13} \times 60 \approx 73.85 \, \text{pF} $.

Question 2

Two capacitors of $ 60 \, \text{pF} $ and $ 120 \, \text{pF} $ are connected in series. What is the equivalent capacitance?

Accepted Answer

$ \frac{1}{C} = \frac{1}{60} + \frac{1}{120} = \frac{2 + 1}{120} = \frac{3}{120} $. $ C = \frac{120}{3} = 40 \, \text{pF} $.

Question 3

A dipole with $ p = 6 \times 10^{-9} \, \text{C m} $ makes an angle of $ 45^\circ $ with a uniform field $ E = 5 \times 10^4 \, \text{N/C} $. What is its potential energy?

Accepted Answer

$ U = -p E \cos \theta = -6 \times 10^{-9} \times 5 \times 10^4 \times \cos 45^\circ $. $ \cos 45^\circ = \frac{1}{\sqrt{2}} \approx 0.707 $, so $ U = -6 \times 10^{-9} \times 5 \times 10^4 \times 0.707 = -2.121 \times 10^{-4} \, \text{J} $.

NEET Physics: Electrostatic Potential and Capacitance — Practice Set 9

Q1. A parallel plate capacitor with capacitance \( 60 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/4 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Q2. Two capacitors of \( 60 \, \text{pF} \) and \( 120 \, \text{pF} \) are connected in series. What is the equivalent capacitance?

Q3. A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 5 \times 10^4 \, \text{N/C} \). What is its potential energy?

Q4. A parallel plate capacitor with capacitance \( 120 \, \text{pF} \) has a dielectric (\( K = 4 \), thickness \( d/4 \)) inserted. What is the new capacitance? (Original separation \( d \)).

Q5. Why does the potential energy of an electric dipole in a uniform electric field become minimum when the dipole aligns with the field?

Q6. A capacitor of \( 5 \, \mu\text{F} \) is charged to \( 100 \, \text{V} \). What is the energy stored in it?

Q7. A point charge \( Q = 24 \times 10^{-9} \, \text{C} \) is placed at the origin. Calculate the potential at a point 8 m away from the charge. (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q8. A spherical conductor of radius 25 cm has a charge of \( 10 \times 10^{-8} \, \text{C} \). What is the electric field at 60 cm from the center? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q9. An electric dipole with moment \( p = 6 \times 10^{-10} \, \text{C m} \) lies along the x-axis. What is the potential at \( (5, 0, 0) \, \text{m} \)? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q10. A dipole with \( p = 3 \times 10^{-10} \, \text{C m} \) makes an angle of \( 90^\circ \) with a uniform field \( E = 2 \times 10^5 \, \text{N/C} \). What is its potential energy?