A charged soap bubble of radius R has surface charge density σ. The excess pressure inside the bubble due to electrostatic force is:
Answer: A
Electrostatic pressure = ε₀E²/2 at surface. E = σ/ε₀ just outside. Excess pressure p = σ²/(2ε₀)
Q.2Hard
A point charge q is placed at distance r from an infinite grounded conducting plane. The force on the charge is:
Answer: A
By method of images, image charge -q is at distance r behind plane. Total distance = 2r. F = kq²/(2r)² = q²/(16πε₀r²)
Q.3Hard
The self-energy of a uniformly charged sphere of radius R and total charge Q is:
Answer: A
Self-energy of uniformly charged sphere: U = 3Q²/(20πε₀R) = 3kQ²/(5R)
Q.4Hard
If potential varies as V = 3x² + 4y in a region, the electric field at point (1,2) is:
Answer: A
E = -∇V = -(∂V/∂x i + ∂V/∂y j) = -(6x i + 4j) = -6i - 4j at (1,2)
Q.5Hard
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:
Answer: A
By symmetry, perpendicular components cancel. Axial component: E = λ/(2πε₀y) × L/√(L²/4 + y²) = λL/(2πε₀y√(L²/4 + y²))
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Q.6Hard
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:
Answer: C
Using potential superposition and the dipole configuration, ΔV = 2kQd(1/x - 21x) for points on the perpendicular bisector.
Q.7Hard
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:
Answer: A
Integrating potential contributions from small elements: V = (kQ/L)ln[(x+L)/x].
Q.8Hard
A uniformly charged infinite line with linear charge density λ creates an electric field at perpendicular distance r. What is E?
Answer: B
Using Gauss's law for infinite line: E = λ/(2πε₀r). This is standard result for line charge.
Q.9Hard
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.10Hard
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.11Hard
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.12Hard
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.13Hard
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.
