RPSC ASSISTANT PROFESSOR PHYSICS MCQ:-
Electrostatics: Gauss’s Law and its Applications
1. Gauss’s law is valid for:
A) Only open surfaces
B) Only closed surfaces
C) Both open and closed surfaces
D) Only planar surfaces
Answer: B
Explanation: Gauss’s law is valid only for closed surfaces, as it relates electric flux through a closed surface to the charge enclosed1.
2. The mathematical form of Gauss’s law is:
A) E⃗⋅A⃗=QE⋅A=Q
B) ΦE=Qenclosedϵ0ΦE=ϵ0Qenclosed
C) ∇⋅E⃗=0∇⋅E=0
D) ∇×E⃗=0∇×E=0
Answer: B
Explanation: The total electric flux through a closed surface is ΦE=Qenclosedϵ0ΦE=ϵ0Qenclosed1.
3. If the radius of a Gaussian surface is doubled, the electric flux through it:
A) Doubles
B) Halves
C) Remains the same
D) Becomes zero
Answer: C
Explanation: Electric flux depends only on the charge enclosed, not the size of the surface1.
4. Gauss’s law can be used to calculate:
A) Electric field due to a point charge
B) Magnetic field due to a steady current
C) Both electric and magnetic fields
D) None of these
Answer: A
Explanation: Gauss’s law is used for electric fields; magnetic fields require Ampere’s law1.
5. The SI unit of electric flux is:
A) V/m
B) Nm²/C
C) C/m²
D) Nm/C
Answer: B
Explanation: Electric flux = Electric field × Area = (N/C) × m² = Nm²/C.
6. The total electric flux through a closed surface is proportional to:
A) Area of the surface
B) Total charge enclosed
C) Distance from the charge
D) Permittivity of free space
Answer: B
Explanation: Flux depends only on the total charge enclosed1.
7. The electric field inside a conductor in electrostatic equilibrium is:
A) Maximum
B) Zero
C) Infinite
D) Equal to the applied field
Answer: B
Explanation: In electrostatic equilibrium, the electric field inside a conductor is zero.
8. The electric field just outside a charged conductor is:
A) Zero
B) Directed parallel to the surface
C) Perpendicular to the surface
D) Randomly oriented
Answer: C
Explanation: The field is perpendicular to the surface of a conductor.
9. The electric field at the surface of a conductor with surface charge density σσ is:
A) σϵ0ϵ0σ
B) σϵ0σϵ0
C) ϵ0σσϵ0
D) σ2ϵ02ϵ0σ
Answer: A
Explanation: E=σϵ0E=ϵ0σ just outside the surface.
10. Which law is consistent with the inverse square dependence seen in Coulomb’s law?
A) Gauss’s law
B) Faraday’s law
C) Ampere’s law
D) Lenz’s law
Answer: A
Explanation: Gauss’s law leads to Coulomb’s law and is consistent with the inverse square law1.
Applications of Gauss’s Law
11. The electric field due to an infinite line of charge at a distance rr is:
A) λ2πϵ0r2πϵ0rλ
B) λ4πϵ0r24πϵ0r2λ
C) λϵ0rϵ0rλ
D) λ2πr2πrλ
Answer: A
Explanation: Derived using a cylindrical Gaussian surface.
12. The electric field outside a uniformly charged spherical shell of radius RR and total charge QQ at distance r>Rr>R is:
A) 00
B) Q4πϵ0r24πϵ0r2Q
C) Q4πϵ0R24πϵ0R2Q
D) Qϵ0rϵ0rQ
Answer: B
Explanation: The field is as if all charge were concentrated at the center.
13. The electric field inside a uniformly charged solid sphere (at distance r<Rr<R) is:
A) 00
B) Q4πϵ0r24πϵ0r2Q
C) Qr4πϵ0R34πϵ0R3Qr
D) Q4πϵ0R24πϵ0R2Q
Answer: C
Explanation: Use Gauss’s law for a sphere of radius rr inside the solid sphere.
14. The electric field inside a cavity within a uniformly charged sphere is:
A) Zero
B) Uniform
C) Varies with position
D) Infinite
Answer: C
Explanation: The field varies depending on the position inside the cavity.
15. The electric field due to an infinite plane sheet of charge with surface density σσ is:
A) σ2ϵ02ϵ0σ
B) σϵ0ϵ0σ
C) σ4ϵ04ϵ0σ
D) 2σϵ0ϵ02σ
Answer: A
Explanation: For an infinite sheet, E=σ2ϵ0E=2ϵ0σ.
16. The electric field between two large, oppositely charged parallel plates (neglecting edge effects) is:
A) σϵ0ϵ0σ
B) σ2ϵ02ϵ0σ
C) 00
D) 2σϵ0ϵ02σ
Answer: A
Explanation: Field doubles as contributions from both plates add up.
17. The flux through one face of a cube with a point charge qq at its center is:
A) q6ϵ06ϵ0q
B) qϵ0ϵ0q
C) q4πϵ04πϵ0q
D) 00
Answer: A
Explanation: Total flux is divided equally among the six faces.
18. The electric field at a point outside a uniformly charged spherical shell is independent of:
A) Distance from the center
B) The total charge
C) The radius of the shell
D) Permittivity
Answer: C
Explanation: The field depends only on total charge and distance from center, not shell radius.
19. The electric field inside a conductor is zero because:
A) Charges are stationary
B) Charges move to cancel any internal field
C) There are no charges
D) None of these
Answer: B
Explanation: Free charges move to cancel any internal field.
20. The net flux through a closed surface not enclosing any charge is:
A) Zero
B) Maximum
C) Minimum
D) Infinite
Answer: A
Explanation: By Gauss’s law, if no charge is enclosed, net flux is zero.
Laplace’s and Poisson’s Equations
21. Laplace’s equation in Cartesian coordinates is:
A) ∇2V=0∇2V=0
B) ∇2V=−ρϵ0∇2V=−ϵ0ρ
C) ∇⋅E⃗=0∇⋅E=0
D) ∇×E⃗=0∇×E=0
Answer: A
Explanation: Laplace’s equation applies in regions with zero charge density2.
22. Poisson’s equation is:
A) ∇2V=0∇2V=0
B) ∇2V=−ρϵ0∇2V=−ϵ0ρ
C) ∇⋅E⃗=0∇⋅E=0
D) ∇×E⃗=0∇×E=0
Answer: B
Explanation: Poisson’s equation applies where charge density ρρ is nonzero3.
23. The Laplacian operator ∇2∇2 in three dimensions is:
A) ∂2∂x2+∂2∂y2+∂2∂z2∂x2∂2+∂y2∂2+∂z2∂2
B) ∂∂x+∂∂y+∂∂z∂x∂+∂y∂+∂z∂
C) ∂2∂x2−∂2∂y2∂x2∂2−∂y2∂2
D) ∂∂x−∂∂y∂x∂−∂y∂
Answer: A
Explanation: This is the definition of the Laplacian in Cartesian coordinates.
24. Laplace’s equation is satisfied in regions where:
A) Charge density is zero
B) Charge density is maximum
C) Potential is constant
D) Electric field is zero
Answer: A
Explanation: Laplace’s equation applies in charge-free regions2.
25. If the potential VV satisfies Laplace’s equation, then:
A) VV is constant everywhere
B) VV can have maxima or minima inside the region
C) VV cannot have a local maximum or minimum inside the region
D) None of these
Answer: C
Explanation: By the maximum-minimum principle, VV cannot have local extrema inside the region.
26. In Poisson’s equation, if ρ=0ρ=0, it reduces to:
A) Gauss’s law
B) Laplace’s equation
C) Faraday’s law
D) Ampere’s law
Answer: B
Explanation: If charge density is zero, Poisson’s equation becomes Laplace’s equation3.
27. The physical meaning of Poisson’s equation is:
A) Relates charge density to electric potential
B) Relates magnetic field to current
C) Relates electric field to magnetic field
D) None of these
Answer: A
Explanation: Poisson’s equation links the spatial variation of potential to charge density3.
28. The general solution to Laplace’s equation depends on:
A) Boundary conditions
B) Initial conditions
C) Both
D) Neither
Answer: A
Explanation: The solution depends on the boundary conditions specified2.
29. For a region with constant charge density, the potential satisfies:
A) Laplace’s equation
B) Poisson’s equation
C) Both
D) None
Answer: B
Explanation: Nonzero charge density means Poisson’s equation applies3.
30. The differential form of Gauss’s law is:
A) ∇⋅E⃗=ρϵ0∇⋅E=ϵ0ρ
B) ∇×E⃗=0∇×E=0
C) ∇⋅B⃗=0∇⋅B=0
D) ∇⋅E⃗=0∇⋅E=0
Answer: A
Explanation: This is the differential form of Gauss’s law