NEET Physics: Electric Charges and Fields questions with solutions

Q1. A Gaussian surface in the figure is shown by dotted line. The electric field on the surface will be:

1. due to \( q_{1} \) and \( q_{2} \) only
2. due to \( q_{2} \) only
3. zero
4. due to all

Answer: due to all

The electric field at any point on the Gaussian surface is the vector sum of contributions from all charges in the region. Gauss’s law relates the net flux through the surface to enclosed charge, but it does not mean only enclosed charges create the field on the surface.

Q2. The acceleration of a charged particle in a uniform electric field is : (where specific charge of a particle \( \left.=\frac{q}{m}\right) \)

1. proportional to its charge only
2. inversely proportional to its mass only
3. proportional to its specific charge
4. inversely proportional to specific charge

Answer: proportional to its specific charge

In a uniform electric field, the electric force on a particle is F = qE. Using F = ma gives a = qE/m = (q/m)E, so acceleration depends on the specific charge q/m. Therefore it is proportional to specific charge.

Q3. The unit of permittivity of free space, ε₀, is:

1. Coulomb²/(Newton-metre²)
2. Coulomb/Newton-metre²
3. Newton-metre²/Coulomb²
4. Coulomb²/Newton-metre²

Answer: Coulomb²/(Newton-metre²)

The unit of permittivity of free space, ε₀, is derived from Coulomb's law, where force is proportional to the product of charges divided by the square of the distance. Rearranging the formula gives ε₀'s unit as Coulomb²/(Newton-metre²).

Q4. Two identical charged spheres suspended from a common point by two massless strings of lengths l, are initially at a distance d (d << l) apart because of their mutual repulsion. The charges begin to leak from both the spheres at a constant rate. As a result, the spheres approach each other with a velocity v. Then v varies as a function of the distance x between the spheres, as:

1. v ∝ x²
2. v ∝ x
3. v ∝ 1/x²
4. v ∝ x⁻¹

Answer: v ∝ x⁻¹

As the charges leak, the electrostatic force decreases, causing the spheres to approach each other. The velocity v is proportional to the rate of change of force, which depends on the inverse square of the distance x. Thus, v ∝ x⁻¹.

Q5. Suppose the charge of a proton and an electron differ slightly. One of them is −e, the other is (e + Δe). If the net electrostatic force and gravitational force between two hydrogen atoms placed at a distance d (much greater than atomic size) apart is zero, then Δe is of the order of [Given mass of hydrogen mₕ = 1.67 × 10⁻²⁷ kg]:

1. 10⁻²³ C
2. 10⁻³⁷ C
3. 10⁻⁴⁷ C
4. 10⁻²⁰ C

Answer: 10⁻⁴⁷ C

For the net force to be zero, the electrostatic force must balance the gravitational force. Using Coulomb's law and Newton's law of gravitation, we equate k(Δe)^2/d^2 = G(mₕ)^2/d^2. Simplifying, Δe = √(G * (mₕ)^2 / k). Substituting values, Δe ≈ 10⁻⁴⁷ C.

Q6. Two point charges A and B, having charges +Q and −Q respectively, are placed at certain distance apart and force acting between them is F. If 25% charge of A is transferred to B, then force between the charges becomes:

1. F
2. 9F/16
3. 16F/9
4. 4F/3

Answer: 9F/16

Initially, the force is F = k(Q)(-Q)/r². After transferring 25% of charge from A to B, charges become (3Q/4) and (-3Q/4). The new force is F' = k(3Q/4)(-3Q/4)/r² = (9/16)F.

Q7. Two parallel infinite line charges with linear charge densities +λ C/m and -λ C/m are placed at a distance of 2R in free space. What is the electric field mid-way between the two line charges?

1. zero
2. 2λ / πε₀ R N/C
3. λ / πε₀ R N/C
4. λ / 2πε₀ R N/C

Answer: 2λ / πε₀ R N/C

The electric field due to each line charge at the midpoint is equal in magnitude but opposite in direction. Since the charges are of opposite signs, the fields add up. The magnitude of the electric field due to one line charge at a distance R is λ / (2πε₀R). Adding the contributions from both charges gives the total field as 2λ / (πε₀R).

Q8. An electron falls from rest through a vertical distance h in a uniform and vertically upward directed electric field E. The direction of electric field is now reversed, keeping its magnitude the same. A proton is allowed to fall from rest in it through the same vertical distance h. The time of fall of the electron, in comparison to the time of fall of the proton is

1. smaller
2. 5 times greater
3. equal
4. 10 times greater

Answer: smaller

The time of fall depends on the acceleration, which is inversely proportional to the mass of the particle (a = qE/m). Since the electron has a much smaller mass than the proton, its acceleration will be much greater, leading to a smaller time of fall.

Q9. An electric dipole of dipole moment p⃗ is aligned parallel to a uniform electric field E⃗. The energy required to rotate the dipole by 90° is

1. pE²
2. p²E
3. pE
4. infinity

Answer: pE

The potential energy of a dipole in a uniform electric field is given by U = -pEcosθ. Initially, the dipole is aligned parallel to the field (θ = 0°), so U_initial = -pE. After rotating it by 90° (θ = 90°), U_final = 0. The energy required is the change in potential energy, ΔU = U_final - U_initial = 0 - (-pE) = pE.

Q10. A hollow metal sphere of radius R is uniformly charged. The electric field due to the sphere at a distance r from the centre:

1. increases as r increases for r < R and for r > R
2. zero as r increases for r < R, decreases as r increases for r > R
3. zero as r increases for r < R, increases as r increases for r > R
4. decreases as r increases for r < R and for r > R

Answer: zero as r increases for r < R, decreases as r increases for r > R

Inside a hollow charged sphere (r < R), the electric field is zero due to the spherical symmetry and Gauss's law. Outside the sphere (r > R), the field decreases with distance as it behaves like a point charge, following the inverse square law.

Q11. A charge Q is placed at the corner of a cube. The electric flux through all the six faces of the cube is

1. Q / 3ε₀
2. Q / 6ε₀
3. Q / 8ε₀
4. Q / ε₀

Answer: Q / 8ε₀

When a charge Q is placed at the corner of a cube, it is shared by 8 cubes. By Gauss's law, the total flux due to charge Q is Q/ε₀. Since the charge is shared equally, the flux through one cube is (Q/8ε₀).

Q12. A point charge +q is placed at mid point of a cube of side 'L'. The electric flux emerging from the cube is

1. q / ε₀
2. 6qL² / ε₀
3. q / 6L²ε₀
4. zero

Answer: q / ε₀

According to Gauss's Law, the total electric flux through a closed surface is equal to the net charge enclosed divided by ε₀. Since the charge +q is at the center of the cube, the flux through the entire cube is q/ε₀.

Q13. A charge Q μC is placed at the centre of a cube, the flux coming out from any surface will be

1. Q / 6ε₀ × 10⁻⁶
2. Q / 6ε₀ × 10⁻³
3. Q / 24ε₀
4. Q / 8ε₀

Answer: Q / 6ε₀ × 10⁻⁶

By Gauss's law, the total flux through a closed surface is given by Φ = Q/ε₀. Since the charge is symmetrically placed at the center of the cube, the flux is equally distributed among its six faces. Thus, the flux through one face is Φ_face = Q / (6ε₀). Converting Q from μC to C introduces a factor of 10⁻⁶, giving the result as Q / 6ε₀ × 10⁻⁶.

Q14. A short electric dipole has a dipole moment of 16 × 10⁻⁹ C·m. The electric potential due to the dipole at a point at a distance of 0.6 m from the centre of the dipole, situated on a line making an angle of 60° with the dipole axis is:

1. 200V
2. 400V
3. zero
4. 50V

Answer: 50V

The electric potential due to a dipole at a point is given by V = (1/4πε₀) * (p * cosθ / r²). Substituting p = 16 × 10⁻⁹ C·m, θ = 60°, r = 0.6 m, and ε₀ = 8.85 × 10⁻¹² F/m, we get V = 50V.

Q15. In a certain region of space with volume 0.2 m³, the electric potential is found to be 5 V throughout. The magnitude of electric field in this region is:

1. 0.5 N/C
2. 1 N/C
3. 5 N/C
4. zero

Answer: zero

The electric field is the spatial rate of change of electric potential. Since the potential is constant (5 V) throughout the region, there is no change in potential, and hence the electric field is zero.

Q16. An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F between the two is:

1. K e² / r³ r̂
2. K e² / r² r̂
3. -K e² / r³ r̂
4. -K e² / r² r̂ (where K = 1 / 4πε₀)

Answer: -K e² / r² r̂ (where K = 1 / 4πε₀)

The Coulomb force between the electron and the nucleus is attractive, so it is directed towards the nucleus. The magnitude of the force is given by Coulomb's law, F = K e² / r², and the negative sign indicates the direction of the force is opposite to the radial vector r̂.

Q17. A charge ‘q’ is placed at the centre of the line joining two equal charges ‘Q’. The system of the three charges will be in equilibrium if ‘q’ is equal to

1. Q/2
2. -Q/4
3. Q/4
4. -Q/2

Answer: -Q/4

For the system to be in equilibrium, the net force on the charge 'q' at the center must be zero. The two charges 'Q' exert forces on 'q' in opposite directions. To balance these forces, 'q' must have a magnitude of Q/4 and an opposite sign to 'Q', i.e., q = -Q/4.

Q18. An electric dipole of moment p⃗ is placed in an electric field of intensity E⃗. The dipole acquires a position such that the axis of the dipole makes an angle θ with the direction of the field. Assuming that the potential energy of the dipole to be zero when θ = 90°, the torque and the potential energy of the dipole will respectively be:

1. pE sin θ, −pE cos θ
2. pE sin θ, −2pE cos θ
3. pE sin θ, 2pE cos θ
4. pE cos θ, −pE cos θ

Answer: pE sin θ, −pE cos θ

The torque on a dipole in an electric field is given by τ = pE sin θ, and the potential energy is U = -pE cos θ, with the reference point of zero potential energy at θ = 90°. Thus, the correct answer is option A.

Q19. Point charges +4q, -q and +4q are kept on the X-axis at points x=0, x=a and x=2a respectively. Then

1. only -q is in static equilibrium
2. none of the charges is in equilibrium
3. all the charges are in unstable equilibrium
4. all the charges are in stable equilibrium

Answer: only -q is in static equilibrium

The charge -q is in static equilibrium because the forces due to the two +4q charges on either side of it are equal in magnitude and opposite in direction, canceling each other out. The other charges are not in equilibrium as there is a net force acting on them.

Q20. A spherical conductor of radius 10 cm has a charge of 3.2 × 10^-7 C distributed uniformly. What is the magnitude of electric field at a point 15 cm from the centre of the sphere?

1. 1.28 × 10^5 N/C
2. 1.28 × 10^6 N/C
3. 1.28 × 10^7 N/C
4. 1.28 × 10^4 N/C

Answer: 1.28 × 10^5 N/C

The electric field outside a spherical conductor is given by E = kQ/r², where k = 9 × 10^9 N·m²/C², Q = 3.2 × 10^-7 C, and r = 0.15 m. Substituting these values, E = (9 × 10^9 × 3.2 × 10^-7) / (0.15)² = 1.28 × 10^5 N/C.