A point charge q is placed at distance r from an infinite grounded conducting plane. The force on the charge is:
Answer: A
By method of images, image charge -q is at distance r behind plane. Total distance = 2r. F = kq²/(2r)² = q²/(16πε₀r²)
Q.202Hard
The self-energy of a uniformly charged sphere of radius R and total charge Q is:
Answer: A
Self-energy of uniformly charged sphere: U = 3Q²/(20πε₀R) = 3kQ²/(5R)
Q.203Easy
An electric dipole with dipole moment p is placed in uniform electric field E at angle θ to the field. The torque on dipole is:
Answer: A
Torque on dipole in uniform field: τ = p × E = pE sinθ, where θ is angle between dipole moment and field
Q.204Easy
Two identical conducting spheres carry charges Q₁ and Q₂. They are brought in contact and then separated. The final charge on each sphere is:
Answer: A
When identical conducting spheres touch, charge distributes equally. Final charge on each = (Q₁ + Q₂)/2 by charge conservation
Q.205Easy
The electric field inside a uniformly charged spherical shell of radius R and charge Q at a distance r from center (r < R) is:
Answer: B
By Gauss's law, electric field inside a uniformly charged spherical shell is zero everywhere
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Q.206Medium
A charged particle moves from point A to point B in electrostatic field. The work done by electrostatic force is independent of path because:
Answer: A
Electrostatic force is conservative, meaning work depends only on initial and final positions, not the path taken
Q.207Medium
A conducting sphere of radius R₁ inside a conducting spherical shell of radius R₂ (R₁ < R₂) has charge Q. The electric field for r > R₂ is:
Answer: B
By Gauss's law, field outside depends only on total enclosed charge Q. E = kQ/r² for r > R₂
Q.208Medium
The capacitance of a spherical conductor of radius R is:
Answer: B
Capacitance of isolated sphere: C = 4πε₀R, where R is radius. Since k = 1/(4πε₀), C = R/k
Q.209Hard
If potential varies as V = 3x² + 4y in a region, the electric field at point (1,2) is:
Answer: A
E = -∇V = -(∂V/∂x i + ∂V/∂y j) = -(6x i + 4j) = -6i - 4j at (1,2)
Q.210Medium
The potential energy of electric dipole in uniform field E is:
Answer: A
Potential energy of dipole: U = -p·E = -pE cosθ, where θ is angle between p and E. Minimum at θ = 0
Q.211Easy
A uniformly charged infinite plane sheet has surface charge density σ. What is the electric field at a distance d from the sheet?
Answer: A
For an infinite uniformly charged plane sheet, using Gauss's law with a cylindrical Gaussian surface, E = σ/(2ε₀), independent of distance.
Q.212Medium
Two point charges +8μC and -2μC are separated by 3 m. The electric potential is zero at a point on the line joining them, located at distance x from the +8μC charge. Find x:
Answer: C
For zero potential: 8/(x) = 2/(3-x). Solving: 8(3-x) = 2x → 24 = 10x → x = 2.4 m
Q.213Hard
A charged rod of length L with linear charge density λ is placed along the x-axis. The electric field at a point on the perpendicular bisector at distance y from the center is:
Answer: A
By symmetry, perpendicular components cancel. Axial component: E = λ/(2πε₀y) × L/√(L²/4 + y²) = λL/(2πε₀y√(L²/4 + y²))
Q.214Medium
A parallel plate capacitor with plate separation d and area A is charged to voltage V. A dielectric of dielectric constant K is inserted between the plates. The change in stored energy is:
Answer: A
If isolated (constant charge): U ∝ 1/C, and C increases by K, so U decreases by K. If connected to battery (constant V): U ∝ C, increases by K. Given 'charged' implies isolated.
Q.215Medium
Two conducting spheres of radii r₁ and r₂ (r₁ > r₂) have charges Q₁ and Q₂. They are connected by a wire. The final charge distribution will be such that:
Answer: D
Connected spheres have same potential. V = kQ/r, so Q ∝ r. Both statements are equivalent and correct.
Q.216Medium
The electric field on the axis of a uniformly charged ring of radius a and total charge Q at distance x from the center is:
Answer: A
By symmetry, radial components cancel. Axial component: E = kQx/(a² + x²)^(23)
Q.217Easy
A point charge q is enclosed by a closed surface. If the surface is deformed without changing the charge enclosed, the electric flux through the surface:
Answer: A
By Gauss's law, flux = Q/ε₀, independent of surface shape. Only the enclosed charge matters.
Q.218Easy
An electron is projected horizontally with velocity v into a uniform electric field E pointing downward. Its trajectory is:
Answer: A
Horizontal velocity constant, vertical acceleration due to E-field. Motion similar to projectile motion, hence parabolic.
Q.219Easy
A uniformly charged sphere of radius R and total charge Q has its charge density reduced by half. The electric potential at the surface becomes:
Answer: A
Surface potential V = kQ/R. If charge becomes Q/2, then V becomes kQ/(2R), which is half the original.
Q.220Medium
Two identical metal spheres with charges +Q and -3Q are brought into contact and then separated by distance r. The electrostatic force between them is:
Answer: A
After contact: charge on each = (Q - 3Q)/2 = -Q. Force F = k(-Q)(-Q)/r² = kQ²/r², attractive (opposite signs).