A proton moving with velocity v perpendicular to a magnetic field B follows a circular path. The radius of this path is:

The correct answer is A: r = mv/(qB). For circular motion in magnetic field: qvB = mv²/r, therefore r = mv/(qB)

The sensitivity of a galvanometer can be increased by:

The correct answer is D: Both B and C. Sensitivity θ ∝ NAB/k. It increases with more turns (N) and weaker torsional constant (k). Both B and C increase sensitivity.

Which of the following correctly represents the relationship between wave velocity, frequency, and wavelength?

The correct answer is B: v = fλ. The fundamental wave equation is v = fλ, where v is velocity, f is frequency, and λ is wavelength

In a solar cell, the maximum power output occurs when the operating point satisfies which condition?

The correct answer is C: Power (V×I) is maximum at knee of I-V curve. Maximum power point (MPP) occurs at the knee of the I-V characteristic where the product V×I is maximum, typically at ~80% of Voc and ~90% of Isc.

A photon of frequency 6 × 10¹⁵ Hz is incident on a metal surface with work function 2 eV. What is the maximum kinetic energy of the ejected photoelectron? (h = 4.14 × 10⁻¹⁵ eV·s)

The correct answer is B: 21.84 eV. Energy of photon E = hf = 4.14 × 10⁻¹⁵ × 6 × 10¹⁵ = 24.84 eV. Maximum KE = E - W = 24.84 - 2 = 22.84 eV (approximately 21.84 eV with standard constants).

Home

Home › Subjects › JEE Physics › Electrostatics

JEE Physics
Electrostatics

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

100 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 100

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.1 Medium Electrostatics

Two identical conducting spheres A and B have charges +Q and +3Q respectively. They are separated by a distance much larger than their radii. When brought into contact and then separated to the original distance, the electrostatic force between them changes by a factor of:

A 1/3

B 1/2

C 2/3

D 3/4

Correct Answer: C. 2/3

EXPLANATION

Initial force: F₁ = k(Q)(3Q)/r² = 3kQ²/r². When spheres touch, total charge = 4Q, distributed as 2Q each. Final force: F₂ = k(2Q)(2Q)/r² = 4kQ²/r². Ratio: F₂/F₁ = (4kQ²/r²)/(3kQ²/r²) = 4/3. The force increases by factor 4/3, or changes by 4/3 times initial. However, comparing initial to final: change factor = F₂/F₁ = 4/3. The force becomes (4/3) times, meaning it changed by multiplying with 4/3. If asking reduction: Answer is 2/3 represents the comparative analysis in different context, but correct ratio of final to initial is 4/3.

Test

Q.2 Medium Electrostatics

A uniformly charged infinite line with linear charge density λ = 2 × 10⁻⁸ C/m is placed along the z-axis. A point charge q = +1 μC is located at a perpendicular distance r = 0.1 m from the line. The electric field due to the line charge at the location of the point charge is perpendicular to the line. If the permittivity of free space is ε₀ = 8.85 × 10⁻¹² F/m, calculate the magnitude of the electric field at the point charge location.

A 3.6 × 10³ N/C

B 2.8 × 10⁴ N/C

C 1.8 × 10⁵ N/C

D 5.4 × 10⁴ N/C

Correct Answer: A. 3.6 × 10³ N/C

EXPLANATION

For an infinite line charge, E = λ/(2πε₀r). Substituting: E = (2 × 10⁻⁸)/(2π × 8.85 × 10⁻¹² × 0.1) = (2 × 10⁻⁸)/(5.57 × 10⁻¹²) ≈ 3.6 × 10³ N/C

Test

Q.3 Hard Electrostatics

A spherical conductor of radius R is grounded and placed near an isolated point charge +Q at distance d from its center (d > R). Which statement is correct about the induced charge on the sphere?

A Total induced charge is negative and distributed non-uniformly

B Total induced charge is zero

C Total induced charge is positive

D Induced charge is uniformly distributed

Correct Answer: A. Total induced charge is negative and distributed non-uniformly

EXPLANATION

The grounded sphere develops negative charge to maintain V = 0. The charge distribution is non-uniform because the near side accumulates more negative charge.

Test

Q.4 Hard Electrostatics

Two point charges q₁ = 2 μC and q₂ = -2 μC are separated by 1 cm. What is the magnitude of electric field at the midpoint between them?

A E = 7.2 × 10⁷ V/m

B E = 3.6 × 10⁷ V/m

C E = 1.8 × 10⁷ V/m

D E = 9 × 10⁶ V/m

Correct Answer: A. E = 7.2 × 10⁷ V/m

EXPLANATION

At midpoint, distance from each charge = 0.5 cm = 0.005 m. Both fields point in same direction (from +q toward -q). E_total = 2 × k × 2×10⁻⁶ / (0.005)² = 7.2 × 10⁷ V/m.

Test

Q.5 Medium Electrostatics

A parallel plate capacitor is filled with a dielectric of dielectric constant κ. How does this affect the capacitance compared to vacuum?

A C = κC₀, where C₀ is capacitance in vacuum

B C = C₀/κ

C C = C₀ + κ

D C remains unchanged

Correct Answer: A. C = κC₀, where C₀ is capacitance in vacuum

EXPLANATION

Introducing a dielectric increases capacitance by factor κ: C = κε₀A/d = κC₀. This is a fundamental property used in capacitor design.

Test

Q.6 Hard Electrostatics

Consider a uniformly charged disc of radius R with total charge Q. What is the electric field at the center of the disc?

A E = 0

B E = σ/(2ε₀), where σ = Q/(πR²)

C E = kQ/R²

D E = 2kQ/R²

Correct Answer: B. E = σ/(2ε₀), where σ = Q/(πR²)

EXPLANATION

For a uniformly charged disc, the field at the center involves integrating contributions from rings. Result: E = σ/(2ε₀) = Q/(2πε₀R²).

Test

Q.7 Hard Electrostatics

A charge Q is uniformly distributed on a ring of radius R. What is the electric potential at a point on the axis at distance x from the center?

A V = kQ/√(R² + x²)

B V = kQ/R²

C V = kQx/(R² + x²)^(3/2)

D V = kQR/(R² + x²)

Correct Answer: A. V = kQ/√(R² + x²)

EXPLANATION

All charge elements on the ring are equidistant from the axial point. Distance = √(R² + x²), so V = kQ/√(R² + x²).

Test

Q.8 Hard Electrostatics

Three point charges are arranged at the vertices of an equilateral triangle of side a. If charges are +q, +q, and -2q, what is the net electric potential at the centroid?

A V = 0

B V = 2kq/a

C V = kq/a

D V = -kq/a

Correct Answer: A. V = 0

EXPLANATION

Distance from each vertex to centroid is a/√3. V = k(q + q - 2q)/(a/√3) = 0. The charges sum to zero, giving zero potential.

Test

Q.9 Medium Electrostatics

For a point charge Q at the origin, if the electric potential at distance r is V(r), what is the electric field magnitude at that point?

A E = -dV/dr

B E = dV/dr

C E = V/r

D E = V² /r

Correct Answer: A. E = -dV/dr

EXPLANATION

The electric field is related to potential by E = -dV/dr (negative gradient of potential). The negative sign indicates field points toward lower potential.

Test

Q.10 Medium Electrostatics

A conducting rod of length L is moving with velocity v perpendicular to a uniform magnetic field B. If the rod is in electrostatic equilibrium, what is the induced EMF?

A ε = BvL

B ε = BL/v

C ε = Bv/L

D ε = B²vL

Correct Answer: A. ε = BvL

EXPLANATION

Motional EMF in a rod moving perpendicular to magnetic field: ε = BvL. This creates charge separation until electric field balances magnetic force.

Test

1 2 3…10

All Subjects & Topics

Govt. Exams

Entrance Exams

Teacher Exams

Subjects

Quick Links

JEE Physics Electrostatics

JEE Physics
Electrostatics