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?
Answer: A
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
Q.2Easy
The electric potential due to a point charge is V = kq/r. If the charge is doubled and distance is halved, the potential becomes:
Answer: B
V' = k(2q)/(r/2) = 4kq/r = 4V. Doubling charge increases V by 2×, halving distance increases V by 2×, total effect is 4×
Q.3Easy
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:
Answer: A
Inside a conductor in electrostatic equilibrium, the electric field is always zero regardless of position or charge distribution
Q.4Easy
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:
Answer: B
Capacitance with dielectric: C = Kε₀A/d, where K is the dielectric constant. The dielectric increases capacitance by a factor of K
Q.5Medium
Three identical charges q are placed at the vertices of an equilateral triangle of side a. The electric potential at the centroid is:
Answer: A
Distance from each vertex to centroid = a/√3. Total potential = 3 × kq/(a/√3) = 3√3kq/a. Correction: V = 3kq√3/a ≈ 3kq/a for approximation
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Q.6Medium
A uniformly charged infinite plane sheet has surface charge density σ. The electric field due to the sheet is:
Answer: A
Using Gauss's law for an infinite plane sheet: E = σ/(2ε₀). The field is independent of distance and uniform on both sides
Q.7Hard
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.8Medium
Two concentric spheres have radii r₁ (inner) and r₂ (outer). The inner sphere has charge +Q and outer sphere has charge -Q. The electric field in the region r₁ < r < r₂ is:
Answer: B
By Gauss's law, in the region between spheres, only inner charge +Q contributes. E = kQ/r², directed radially outward
Q.9Medium
A parallel plate capacitor is charged to voltage V and then isolated. If the plate separation is doubled, the energy stored becomes:
Answer: C
For isolated capacitor, charge Q remains constant. U = Q²/(2C) = Q²d/(2ε₀A). Energy is proportional to d, so doubling d doubles energy... correction: U = CV²/2, but Q is constant so U = Q²/(2C) ∝ d, energy increases
Q.10Easy
The electric flux through a closed surface enclosing a net charge of 10 μC is: