NEET Physics: Electrostatic Potential and Capacitance — Practice Set 12

Q1. A spherical conductor of radius 4 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the potential at its surface? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).

Q2. A conductor has a surface charge density of \( 3.5 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

Q3. In a charged parallel plate capacitor, why does the electric field remain uniform between the plates even when a dielectric slab is partially inserted?

Q4. A charged particle moves along a path between two points in an electric field where the potential difference is zero. What can be inferred about the path taken?

Q5. Why does the electric field just outside a charged conductor depend only on the surface charge density and not on the total charge of the conductor?

Q6. Why does the electric field between two large parallel plates with opposite charges remain uniform even if one plate has a dielectric coating on its inner surface?

Q7. Why does the electric field inside a dielectric material decrease when placed in an external field, compared to the field in free space?

Q8. A dipole with \( p = 6 \times 10^{-9} \, \text{C m} \) makes an angle of \( 45^\circ \) with a uniform field \( E = 2 \times 10^5 \, \text{N/C} \). What is its potential energy?

Q9. A conductor has a surface charge density of \( 1 \times 10^{-6} \, \text{C/m}^2 \). What is the electric field just outside it? (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 \text{N}^{-1} \text{m}^{-2} \)).

Q10. A \( 2 \, \mu\text{F} \) capacitor is charged to \( 500 \, \text{V} \). What is the energy stored in it?

PhysicsElectrostatic Potential and Capacitance

Set 12 of 25

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A spherical conductor of radius 4 cm has a charge of \( 4 \times 10^{-8} \, \text{C} \). What is the potential at its surface? (Take \( \frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2} \)).