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JEE Physics

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

434 Q 9 Topics Take Test
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Difficulty: All Easy Medium Hard 291–300 of 434
Topics in JEE Physics
Q.291 Medium Electrostatics
A non-conducting infinite plane with uniform surface charge density σ produces an electric field. What is the magnitude of this field?
A E = σ/(2ε₀)
B E = σ/ε₀
C E = σ²/(2ε₀)
D E = 2σ/ε₀
Correct Answer:  A. E = σ/(2ε₀)
EXPLANATION

Using Gauss's law for an infinite plane: E = σ/(2ε₀). The field is independent of distance and perpendicular to the plane.

Test
Q.292 Medium Electrostatics
A spherical shell of radius R carries a uniformly distributed charge Q. What is the electric field inside the shell at distance r from the center (r < R)?
A E = 0
B E = kQ/r²
C E = kQ/R²
D E = kQr/R³
Correct Answer:  A. E = 0
EXPLANATION

By Gauss's law, for a uniformly charged spherical shell, the electric field inside (r < R) is zero because the enclosed charge is zero.

Test
Q.293 Medium Electrostatics
Two identical conducting spheres have charges +Q and -Q respectively. They are brought into contact and then separated. What is the final charge on each sphere?
A Zero on both
B +Q/2 and -Q/2
C +Q and -Q
D Cannot be determined
Correct Answer:  A. Zero on both
EXPLANATION

When identical conducting spheres touch, charge distributes equally. Total charge = +Q + (-Q) = 0, so each gets 0. They remain neutral after separation.

Test
Q.294 Medium Electrostatics
An electron is released from rest in a uniform electric field of magnitude E. After moving through a distance d, what is its kinetic energy?
A K = eEd
B K = eE/d
C K = eEd²
D K = Ed/e
Correct Answer:  A. K = eEd
EXPLANATION

Work done by electric field = Change in kinetic energy. W = qEd = eEd, which equals kinetic energy since initial KE = 0.

Test
Q.295 Medium Electrostatics
A conducting sphere of radius a is charged to potential V₀. What is the surface charge density on the sphere?
A σ = V₀/(ka)
B σ = kV₀a
C σ = V₀a/k
D σ = kV₀/a
Correct Answer:  A. σ = V₀/(ka)
EXPLANATION

For a conducting sphere, V₀ = kQ/a = kσ(4πa²)/a, which gives σ = V₀/(ka).

Test
Q.296 Medium Electrostatics
Three capacitors of capacitance 2 μF, 3 μF, and 6 μF are connected in series across a 12 V supply. What is the charge on the 2 μF capacitor?
A 4 μC
B 6 μC
C 8 μC
D 12 μC
Correct Answer:  D. 12 μC
EXPLANATION

In series, 1/C_eq = 1/2 + 1/3 + 1/6 = 1. C_eq = 1 μF. Q = C_eq × V = 1 × 12 = 12 μC. Same charge on all capacitors.

Test
Q.297 Medium Electrostatics
A charged particle of mass m and charge q is projected perpendicular to a uniform electric field E. What is the trajectory?
A Circular
B Parabolic
C Elliptical
D Straight line
Correct Answer:  B. Parabolic
EXPLANATION

Perpendicular to field: uniform motion. Parallel to field: constant acceleration. Combined motion is parabolic (similar to projectile motion).

Test
Q.298 Medium Electrostatics
Two point charges q₁ and q₂ are separated by distance r. If they are moved to distance 2r apart, how does the interaction energy change?
A Becomes half
B Remains same
C Doubles
D Becomes four times
Correct Answer:  A. Becomes half
EXPLANATION

U = kq₁q₂/r. When distance doubles, U becomes kq₁q₂/2r = U/2

Test
Q.299 Medium Electrostatics
Two parallel conducting plates have surface charge density σ and -σ respectively. What is the electric field in the region between the plates?
A σ/2ε₀
B σ/ε₀
C σ/4ε₀
D 2σ/ε₀
Correct Answer:  B. σ/ε₀
EXPLANATION

Field due to each plate is σ/2ε₀. Between opposite plates, fields add: σ/2ε₀ + σ/2ε₀ = σ/ε₀

Test
Q.300 Medium Electrostatics
A charge distribution creates potential V(r) = kr² where k is constant. What is the electric field at distance r?
A -2kr
B 2kr
C kr
D -kr²
Correct Answer:  A. -2kr
EXPLANATION

E = -dV/dr = -d(kr²)/dr = -2kr. Field is negative indicating direction opposite to increasing r.

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