Physics is where JEE ranks are usually won or lost, since a single conceptual slip changes the whole answer. Questions here span mechanics, rotational motion, thermodynamics, waves and oscillations, electrostatics, current electricity, magnetism, optics, and modern physics. Numerical problems carry full derivations so you can see which step you skipped, not just which option was right.
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
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:
