A uniformly charged ring of radius R carries total charge Q. What is the electric field at a point on the axis at distance x from the center?

The correct answer is A: kQx/(x² + R²)^(3/2). Using Coulomb's law and symmetry, the radial components cancel. The axial component gives E = kQx/(x² + R²)^(3/2).

An electron in excited state of hydrogen atom has angular momentum 2ℏ. The principal quantum number n is:

The correct answer is C: 3. Angular momentum L = ℏ√(l(l+1)) = 2ℏ gives l(l+1) = 4, so l = 2. Since l < n, minimum n = 3.

In a cathode ray tube with accelerating potential V, electrons reach the anode with kinetic energy. If V is doubled, the maximum frequency of X-rays produced will:

The correct answer is B: Double. Maximum X-ray frequency: fmax = eV/h. If V is doubled, fmax also doubles, since f ∝ V.

The ratio of specific charge (e/m) for an α-particle to that of a proton is:

The correct answer is A: 1/2. α-particle: charge = 2e, mass = 4u. Proton: charge = e, mass = u. (e/m)ₐ/(e/m)ₚ = (2e/4u)/(e/u) = 1/2

A block is pushed up an incline of angle 30° with initial velocity 10 m/s. If coefficient of kinetic friction is 0.2, how far will it travel before stopping? (g = 10 m/s²)

The correct answer is A: 3.2 m. Deceleration a = g(sinθ + μcosθ) = 10(0.5 + 0.2×0.866) = 10(0.5 + 0.173) = 6.73 m/s². Using v² = u² - 2as: 0 = 100 - 2(6.73)s, s ≈ 3.2 m

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JEE Physics
Electrostatics

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

13 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 1–10 of 13

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 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.2 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.3 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.4 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.5 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.6 Hard Electrostatics

A uniformly charged infinite line with linear charge density λ creates an electric field at perpendicular distance r. What is E?

A kλ/r²

B λ/2πε₀r

C λ/πε₀r²

D 2kλ/r

Correct Answer: B. λ/2πε₀r

EXPLANATION

Using Gauss's law for infinite line: E = λ/(2πε₀r). This is standard result for line charge.

Test

Q.7 Hard Electrostatics

An insulating rod of length L is uniformly charged with total charge Q. The electric potential at a point on the axis at distance x from one end is:

A kQ ln[(x+L)/x]/L

B kQ/L

C kQ/(x+L)

D kQ ln(L/x)/L

Correct Answer: A. kQ ln[(x+L)/x]/L

EXPLANATION

Integrating potential contributions from small elements: V = (kQ/L)ln[(x+L)/x].

Test

Q.8 Hard Electrostatics

Two point charges Q and -Q are at distance 2d apart. The potential difference between two points on the perpendicular bisector at distances x and 2x from the midpoint is proportional to:

A Q/d

B Q(1/x - 1/2x)

C Qd(1/x - 1/2x)

D Cannot be determined without more data

Correct Answer: C. Qd(1/x - 1/2x)

EXPLANATION

Using potential superposition and the dipole configuration, ΔV = 2kQd(1/x - 1/2x) for points on the perpendicular bisector.

Test

Q.9 Hard Electrostatics

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:

A λL/(2πε₀y√(L²/4 + y²))

B λ/(4πε₀y)

C λL/(4πε₀(L²/4 + y²))

D λL/(πε₀y²)

Correct Answer: A. λL/(2πε₀y√(L²/4 + y²))

EXPLANATION

By symmetry, perpendicular components cancel. Axial component: E = λ/(2πε₀y) × L/√(L²/4 + y²) = λL/(2πε₀y√(L²/4 + y²))

Test

Q.10 Hard Electrostatics

If potential varies as V = 3x² + 4y in a region, the electric field at point (1,2) is:

A E = -6i - 4j N/C

B E = 6i + 4j N/C

C E = -6i + 4j N/C

D E = 6i - 4j N/C

Correct Answer: A. E = -6i - 4j N/C

EXPLANATION

E = -∇V = -(∂V/∂x i + ∂V/∂y j) = -(6x i + 4j) = -6i - 4j at (1,2)

Test

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