A uniformly charged conducting sphere of radius R and total charge Q is surrounded by a concentric spherical shell. Using Gauss's law, what is the electric field at a distance r > R from the center?
Answer: A
By Gauss's law, for r > R, the electric field depends only on the enclosed charge Q and is E = kQ/r², independent of the outer shell.
Q.82Easy
The electric potential energy of a system of two point charges q₁ and q₂ separated by distance r is:
Answer: A
Potential energy of two point charges is U = kq₁q₂/r. This is positive for like charges (repulsive) and negative for unlike charges (attractive).
Q.83Medium
A conducting sphere of radius a is charged to potential V₀. What is the surface charge density on the sphere?
Answer: A
For a conducting sphere, V₀ = kQ/a = kσ(4πa²)/a, which gives σ = V₀/(ka).
Q.84Medium
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?
Answer: A
Work done by electric field = Change in kinetic energy. W = qEd = eEd, which equals kinetic energy since initial KE = 0.
Q.85Medium
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?
Answer: A
When identical conducting spheres touch, charge distributes equally. Total charge = +Q + (-Q) = 0, so each gets 0. They remain neutral after separation.
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Q.86Medium
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)?
Answer: A
By Gauss's law, for a uniformly charged spherical shell, the electric field inside (r < R) is zero because the enclosed charge is zero.
Q.87Easy
An electric field E = 100 V/m is applied between two parallel plates separated by distance d = 0.02 m. What is the potential difference between the plates?
Answer: A
For uniform field between parallel plates: V = Ed = 100 × 0.02 = 2 V.
Q.88Medium
A non-conducting infinite plane with uniform surface charge density σ produces an electric field. What is the magnitude of this field?
Answer: A
Using Gauss's law for an infinite plane: E = σ/(2ε₀). The field is independent of distance and perpendicular to the plane.
Q.89Medium
A charge +q is at position (0, 0) and charge -q is at (a, 0). At what point on the x-axis is the electric potential zero?
Answer: A
At point (x, 0): V = kq/x - kq/(a-x) = 0 gives x = a-x, so x = a/2. The midpoint has zero potential.
Q.90Medium
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?
Answer: A
Motional EMF in a rod moving perpendicular to magnetic field: ε = BvL. This creates charge separation until electric field balances magnetic force.
Q.91Medium
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?
Answer: A
The electric field is related to potential by E = -dV/dr (negative gradient of potential). The negative sign indicates field points toward lower potential.
Q.92Hard
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?
Answer: A
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.
Q.93Hard
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?
Answer: A
All charge elements on the ring are equidistant from the axial point. Distance = √(R² + x²), so V = kQ/√(R² + x²).
Q.94Hard
Consider a uniformly charged disc of radius R with total charge Q. What is the electric field at the center of the disc?
Answer: B
For a uniformly charged disc, the field at the center involves integrating contributions from rings. Result: E = σ/(2ε₀) = Q/(2πε₀R²).
Q.95Medium
A parallel plate capacitor is filled with a dielectric of dielectric constant κ. How does this affect the capacitance compared to vacuum?
Answer: A
Introducing a dielectric increases capacitance by factor κ: C = κε₀A/d = κC₀. This is a fundamental property used in capacitor design.
Q.96Hard
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?
Answer: A
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.
Q.97Hard
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?
Answer: A
The grounded sphere develops negative charge to maintain V = 0. The charge distribution is non-uniform because the near side accumulates more negative charge.
Q.98Medium
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.
Answer: A
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
Q.99Medium
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:
Answer: C
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) = 34. The force increases by factor 34, or changes by 34 times initial. However, comparing initial to final: change factor = F₂/F₁ = 34. The force becomes (34) times, meaning it changed by multiplying with 34. If asking reduction: Answer is 32 represents the comparative analysis in different context, but correct ratio of final to initial is 34.
