A parallel plate capacitor has plate area A, separation d, and dielectric constant K. Its capacitance is:

The correct answer is B: Kε₀A/d. The capacitance of a parallel plate capacitor with dielectric is C = Kε₀A/d, where K is the dielectric constant.

An electron transitions from n=3 to n=1 in a hydrogen atom. The frequency of emitted photon is (Use Rydberg constant R = 1.097×10⁷ m⁻¹):

The correct answer is A: 2.93×10¹⁵ Hz. Using f = cR(1/n₁² - 1/n₂²) = 3×10⁸ × 1.097×10⁷ × (1/1 - 1/9) = 3×10⁸ × 1.097×10⁷ × (8/9) ≈ 2.93×10¹⁵ Hz

An electron in the first excited state of hydrogen atom (n=2) transitions to ground state (n=1). The wavelength of emitted photon is approximately:

The correct answer is A: 121.6 nm. Using Rydberg formula: 1/λ = R(1/1² - 1/2²) = R(3/4). With R = 1.097 × 10⁷ m⁻¹, λ ≈ 121.6 nm (Lyman alpha line).

The resolving power of a microscope depends on:

The correct answer is B: Wavelength of light and numerical aperture. Resolving power = 1/(1.22λ/2NA). It depends on wavelength and numerical aperture (NA = n×sin(θ))

A body of mass 2 kg moves in a circular path of radius 0.5 m with constant speed. If the centripetal acceleration is 8 m/s², what is the centripetal force required?

The correct answer is A: 16 N. Centripetal force F = ma_c = 2 × 8 = 16 N. This is the net force directed toward the center of circular motion.

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JEE Physics
Electrostatics

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

37 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 31–37 of 37

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.31 Easy Electrostatics

Two identical conducting spheres carry charges Q₁ and Q₂. They are brought in contact and then separated. The final charge on each sphere is:

A (Q₁ + Q₂)/2

B (Q₁ - Q₂)/2

C (Q₁ × Q₂)/2

D Q₁ + Q₂

Correct Answer: A. (Q₁ + Q₂)/2

EXPLANATION

When identical conducting spheres touch, charge distributes equally. Final charge on each = (Q₁ + Q₂)/2 by charge conservation

Test

Q.32 Easy Electrostatics

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:

A τ = pE sinθ

B τ = pE cosθ

C τ = pE tanθ

D τ = pE/sinθ

Correct Answer: A. τ = pE sinθ

EXPLANATION

Torque on dipole in uniform field: τ = p × E = pE sinθ, where θ is angle between dipole moment and field

Test

Q.33 Easy Electrostatics

The electric flux through a closed surface enclosing a net charge of 10 μC is:

A 1.13 × 10⁶ N·m²/C

B 8.85 × 10⁵ N·m²/C

C 2.26 × 10⁶ N·m²/C

D 1.13 × 10⁵ N·m²/C

Correct Answer: A. 1.13 × 10⁶ N·m²/C

EXPLANATION

By Gauss's law: Φ = Q/ε₀ = 10 × 10⁻⁶ / (8.85 × 10⁻¹²) = 1.13 × 10⁶ N·m²/C

Test

Q.34 Easy Electrostatics

A parallel plate capacitor has plates of area A separated by distance d. If a dielectric of constant K is inserted between the plates, the capacitance becomes:

A C = ε₀A/d

B C = Kε₀A/d

C C = ε₀A/(Kd)

D C = ε₀d/A

Correct Answer: B. C = Kε₀A/d

EXPLANATION

Capacitance with dielectric: C = Kε₀A/d, where K is the dielectric constant. The dielectric increases capacitance by a factor of K

Test

Q.35 Easy Electrostatics

A conducting sphere of radius 10 cm carries a charge of 5 μC. The electric field at a distance of 5 cm from the center inside the conductor is:

A 0 N/C

B 4.5 × 10⁵ N/C

C 9 × 10⁵ N/C

D 1.8 × 10⁶ N/C

Correct Answer: A. 0 N/C

EXPLANATION

Inside a conductor in electrostatic equilibrium, the electric field is always zero regardless of position or charge distribution

Test

Q.36 Easy Electrostatics

The electric potential due to a point charge is V = kq/r. If the charge is doubled and distance is halved, the potential becomes:

A 2V

B 4V

C V/2

D V/4

Correct Answer: B. 4V

EXPLANATION

V' = k(2q)/(r/2) = 4kq/r = 4V. Doubling charge increases V by 2×, halving distance increases V by 2×, total effect is 4×

Test

Q.37 Easy Electrostatics

Two point charges of +2 μC and -2 μC are placed 10 cm apart. What is the electric field at a point midway between them?

A 7.2 × 10⁶ N/C

B 3.6 × 10⁶ N/C

C 1.8 × 10⁶ N/C

D 0 N/C

Correct Answer: A. 7.2 × 10⁶ N/C

EXPLANATION

For a dipole configuration, the electric field at the midpoint is E = 2kq/r² directed from negative to positive charge. E = 2 × 9 × 10⁹ × 2 × 10⁻⁶ / (0.05)² = 7.2 × 10⁶ N/C

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Electrostatics