NEET Electric Charges and Fields Previous Year Questions with Solutions

By Coulomb’s law, the initial repulsive force between the two identical spheres A and B carrying charge q separated by distance r is:

• • Spheres are treated as point charges because radii are negligible compared to distance r.

An identical uncharged conducting sphere C (initial charge = 0) is brought into physical contact with sphere A (charge q). Due to identical capacitance and symmetry, charge redistributes equally:

• • Sphere A now retains charge q/2, and sphere C carries charge q/2.

Sphere C (charge q/2) is now brought into contact with sphere B (charge q). The total charge divides equally between the two identical spheres:

• • Sphere C is now removed.
• • The new force between A (charge q/2) and B (charge 3q/4) is:
• • F' = k (q/2)(3q/4) / r² = (3/8) [k q² / r²] = (3/8) F.
• • Corresponds uniquely to Option (4).

The electrostatic potential energy U of an electric dipole with dipole moment p in a uniform electric field E is given by:

• • Stable equilibrium occurs at θ = 0° where U is minimum: U(0°) = -p E.

Evaluate U at initial angle θ₁ = 0° and final angle θ₂ = 60°:

• • Change in potential energy is defined as ΔU = U₂ - U₁.

Calculate the difference and substitute p and E:

• • Matches Option (2) with complete precision.

The electrostatic potential due to a short electric dipole of dipole moment P at a distance r along its axial line is given by:

Substitute 1/(4πε₀) = 9 × 10⁹ N m²/C², P = 4 × 10⁻⁶ C m, and r = 2 m:

The calculated potential matches ±9 × 10³ V exactly. Therefore, Assertion A is completely TRUE.

Reason R claims that the potential formula is V = ± (2P) / (4πε₀ r²). This is scientifically incorrect. The factor of 2 appears in the axial electric field formula E = 2P / (4πε₀ r³), but NOT in the potential formula V = P / (4πε₀ r²). Therefore, Reason R is FALSE.

When an electric dipole of dipole moment p⃗ = q(2a⃗) is placed in a uniform electric field E⃗ at an angle θ, it experiences a restoring torque given by the vector cross product:

Substitute torque τ = 4 N m, length 2a = 2 × 10⁻² m, electric field E = 2 × 10⁵ N/C, and angle θ = 30°:

Evaluate the trigonometric and metric terms using sin 30° = 1/2:

Solve for charge q and express in millicoulombs (1 mC = 10⁻³ C):

The electric flux Φ through a closed Gaussian surface S is geometrically defined as the net count of electric field lines penetrating the surface. Inward-directed lines contribute negative flux, while outward-directed lines contribute positive flux:

The problem states that the number of field lines entering the closed surface equals the number of field lines leaving it. Therefore:

According to Gauss’s Law of electrostatics, the net electric flux through any closed Gaussian surface in vacuum is directly proportional to the total net charge q_enc enclosed within the surface:

Equating the net flux to zero yields:

The electric field lines diverge from a point to the left towards the right. By Faraday's field line principles, field intensity is proportional to the density of lines of force. Therefore, the field is stronger at the position of $+q$ than at the position of $-q$:

Determine the individual forces on each charge: - Force on $+q$: $F_1 = q E_{(+q)}$ acting towards the right (in the direction of $\vec{E}$). - Force on $-q$: $F_2 = q E_{(-q)}$ acting towards the left (opposite to $\vec{E}$).

For any conservative electrostatic system, the net force is the negative gradient of potential energy ($F = -dU/dx$). Any body accelerating in the direction of the net force moves toward a state of lower potential energy: