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JEE Physics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

301 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 221–230 of 301
Topics in JEE Physics
Q.221 Easy Electrostatics
A charged particle enters a uniform electric field perpendicularly. Its path is:
A Straight line
B Parabolic
C Circular
D Hyperbolic
Correct Answer:  B. Parabolic
EXPLANATION

Similar to projectile motion, a charged particle in a perpendicular electric field experiences constant acceleration, resulting in a parabolic trajectory.

Test
Q.222 Easy Electrostatics
The potential at distance r from a point charge Q is V. If the distance is doubled, the new potential is:
A V/2
B 2V
C V/4
D 4V
Correct Answer:  A. V/2
EXPLANATION

Electric potential V = kQ/r is inversely proportional to distance. If r → 2r, then V → V/2.

Test
Q.223 Easy Electrostatics
An electric dipole with dipole moment p is placed in a non-uniform electric field. Which statement is correct?
A The dipole experiences only torque, no net force
B The dipole experiences only force, no torque
C The dipole experiences both torque and net force
D The dipole experiences neither force nor torque
Correct Answer:  C. The dipole experiences both torque and net force
EXPLANATION

In a non-uniform electric field, a dipole experiences both: (1) a torque due to the field trying to align it, and (2) a net force because the fields at the two ends are different.

Test
Q.224 Easy Electrostatics
A uniformly charged infinite plane sheet has surface charge density σ. What is the electric field at a distance from the sheet?
A σ/(2ε₀)
B σ/ε₀
C σ/(4πε₀)
D 2σ/ε₀
Correct Answer:  A. σ/(2ε₀)
EXPLANATION

For an infinite uniformly charged plane sheet, using Gauss's law with a cylindrical Gaussian surface, the electric field is E = σ/(2ε₀), independent of distance from the sheet.

Test
Q.225 Easy Electrostatics
A test charge q is moved from point A to point B in an electric field. The work done by the field is 100 J. If the same charge is moved from B to A, the work done is:
A -100 J
B 100 J
C 0 J
D 200 J
Correct Answer:  A. -100 J
EXPLANATION

Electric force is conservative. Work done in reverse path = negative of original work. W(B→A) = -100 J

Test
Q.226 Easy Electrostatics
An electric dipole experiences maximum torque when placed in a uniform electric field at angle θ with respect to the field. The value of θ is:
A
B 45°
C 90°
D 180°
Correct Answer:  C. 90°
EXPLANATION

Torque τ = pE sin(θ). Maximum when sin(θ) = 1, i.e., θ = 90°

Test
Q.227 Easy Electrostatics
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:
A Half of original
B Same as original
C Double of original
D Quarter of original
Correct Answer:  A. Half of original
EXPLANATION

Surface potential V = kQ/R. If charge becomes Q/2, then V becomes kQ/(2R), which is half the original.

Test
Q.228 Easy Electrostatics
An electron is projected horizontally with velocity v into a uniform electric field E pointing downward. Its trajectory is:
A Parabolic
B Circular
C Straight line
D Elliptical
Correct Answer:  A. Parabolic
EXPLANATION

Horizontal velocity constant, vertical acceleration due to E-field. Motion similar to projectile motion, hence parabolic.

Test
Q.229 Easy Electrostatics
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:
A Remains constant
B Increases
C Decreases
D Becomes zero
Correct Answer:  A. Remains constant
EXPLANATION

By Gauss's law, flux = Q/ε₀, independent of surface shape. Only the enclosed charge matters.

Test
Q.230 Easy Electrostatics
A uniformly charged infinite plane sheet has surface charge density σ. What is the electric field at a distance d from the sheet?
A σ/(2ε₀)
B σ/ε₀
C σ/(4πε₀d²)
D σd/ε₀
Correct Answer:  A. σ/(2ε₀)
EXPLANATION

For an infinite uniformly charged plane sheet, using Gauss's law with a cylindrical Gaussian surface, E = σ/(2ε₀), independent of distance.

Test
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