Questions & explanations
1. Which of the following statements about equipotential surfaces is correct?
- Electric field lines are always parallel to equipotential surfaces
- Work done in moving a charge along an equipotential surface is zero
- Equipotential surfaces are always spherical for a uniform electric field
- The potential difference between two different equipotential surfaces can be zero
Answer: Work done in moving a charge along an equipotential surface is zero
Since every point on an equipotential surface has the same potential, the potential difference ΔV = 0, so work done W = qΔV = 0. Option A is false: field lines are perpendicular to equipotential surfaces. Option C is false: for a uniform field, equipotentials are planes, not spheres. Option D is false: by definition, different equipotential surfaces have different potentials, so ΔV ≠ 0.
2. The electric field at a point is defined as:
- The force experienced by a unit positive charge placed at that point.
- The force experienced by a unit negative charge placed at that point.
- The force experienced by a unit charge irrespective of its sign.
- The force per unit charge on a test charge, provided the test charge is infinitesimally small so as not to disturb the source charges.
Answer: The force per unit charge on a test charge, provided the test charge is infinitesimally small so as not to disturb the source charges.
The precise definition of electric field is E = lim_{q0→0} F/q0, where q0 is a positive test charge. This ensures that the field is the force per unit charge that would be experienced by a small positive test charge. Option A is a simplified version but lacks the condition of small test charge; B and C are incorrect because the test charge is conventionally positive.
3. Which of the following correctly represents Coulomb's law in vector form for the force exerted on charge q1 by charge q2, if r̂12 is the unit vector pointing from q2 to q1?
- F12 = k q1 q2 / r^2 * r̂21
- F12 = k q1 q2 / r^2 * r̂12
- F12 = k q1 q2 / r * r̂12
- F12 = k q1 q2 / r^2 * (-r̂12)
Answer: F12 = k q1 q2 / r^2 * r̂12
Coulomb's law in vector form states that the force on q1 due to q2 is F12 = k q1 q2 / r^2 * r̂12, where r̂12 is the unit vector from q2 to q1. This gives the correct direction (repulsive for like charges, attractive for opposite). Option A uses r̂21 which gives opposite direction; C has r instead of r^2; D has a negative sign making it opposite.
4. Which of the following correctly defines electric potential at a point in an electric field?
- Work done in bringing a unit positive charge from infinity to that point
- Work done in bringing a unit positive charge from that point to infinity
- Force experienced by a unit positive charge placed at that point
- Work done in moving a unit positive charge between any two points in the field
Answer: Work done in bringing a unit positive charge from infinity to that point
Electric potential at a point is defined as the work done per unit charge in bringing a test positive charge from infinity to that point, without acceleration. Option B describes the negative of potential (since potential at infinity is zero), option C defines electric field, and option D defines potential difference, not potential at a point.
5. A solid metallic sphere is placed in a uniform external electric field. After electrostatic equilibrium is reached, what is the electric field inside the sphere?
- Zero everywhere
- Uniform and equal to the external field
- Non-zero but weaker than the external field
- Zero only at the centre but non-zero elsewhere
Answer: Zero everywhere
In electrostatic equilibrium, the free electrons inside a conductor redistribute themselves until the net electric field inside the conductor becomes zero. This is because any internal field would cause further movement of charges until equilibrium is established. Hence, the field inside the entire volume of the conductor is exactly zero.
6. Two point charges +Q and -Q are placed at points A and B separated by a distance 2d. What is the electric field at a point P on the perpendicular bisector of AB at a distance x from the midpoint?
- (kQ d)/( (d^2+x^2)^(3/2) ) directed along the bisector from +Q to -Q.
- (2kQ d)/( (d^2+x^2)^(3/2) ) directed along the bisector from +Q to -Q.
- (kQ x)/( (d^2+x^2)^(3/2) ) directed perpendicular to AB.
- Zero.
Answer: (2kQ d)/( (d^2+x^2)^(3/2) ) directed along the bisector from +Q to -Q.
The fields due to +Q and -Q at P have equal horizontal components along AB and opposite vertical components. The horizontal components add, giving net field E = 2 × (kQ/(d^2+x^2)) × (d/√(d^2+x^2)) = 2kQd/(d^2+x^2)^(3/2) directed from +Q to -Q (parallel to AB). Options A, C, D are incorrect magnitudes or directions.
7. Three point charges +q, +q, -q are placed at the vertices A, B, C respectively of an equilateral triangle of side L. What is the net electrostatic force on the charge at C?
- √3 k q^2 / L^2 directed along the median from C towards the midpoint of AB.
- k q^2 / L^2 directed along the median from C away from the midpoint of AB.
- 2k q^2 / L^2 directed towards vertex A.
- Zero.
Answer: √3 k q^2 / L^2 directed along the median from C towards the midpoint of AB.
The forces on -q at C due to +q at A and +q at B are both attractive, each of magnitude k q^2/L^2, directed along CA and CB respectively. The angle between these forces is 60°. Their resultant has magnitude √3 k q^2/L^2 (since 2F cos30° = √3 F) and is directed along the median from C towards the midpoint of AB.
8. Electrostatic shielding is used to protect sensitive instruments from external electric fields. Which of the following statements best explains the principle behind electrostatic shielding?
- The electric field inside a conductor in electrostatic equilibrium is zero
- The electric field inside a conductor is maximum at its surface
- The electric field inside a conductor is directed radially outward
- The electric field inside a conductor is inversely proportional to the distance from its centre
Answer: The electric field inside a conductor in electrostatic equilibrium is zero
Electrostatic shielding works because when a conductor is placed in an external electric field, charges redistribute to cancel the field inside the conductor material. This ensures that any cavity (and the region inside) is free of electric field, protecting sensitive devices from external electric influences.
9. A charge q is moved from point A to point B on an equipotential surface. The work done by the electric field is:
- qV
- q(V_B - V_A)
- zero
- infinite
Answer: zero
On an equipotential surface, V_A = V_B, so ΔV = 0. Work done by the electric field is W = q(V_A - V_B) = q × 0 = 0. Option A is not a specific expression; option B equals zero but the question asks for the work, and the simplest correct answer is zero. Option D is incorrect.
10. A charge of 2 μC is moved from point A to point B in an electric field. The potential at A is 10 V and at B is 50 V. The work done by the external agent in moving the charge slowly is:
- 80 μJ
- -80 μJ
- 120 μJ
- -120 μJ
Answer: 80 μJ
Work done by an external agent against the electric field equals the change in potential energy: W_ext = q (V_B - V_A) = (2×10⁻⁶ C)(50 V - 10 V) = 2×10⁻⁶ × 40 = 80×10⁻⁶ J = 80 μJ. Option B would be the work done by the electric field. Options C and D are incorrect values.
11. An isolated parallel plate capacitor is charged and then a dielectric slab of dielectric constant K is inserted completely between the plates. Which of the following quantities will increase?
- The charge on the plates
- The capacitance
- The potential difference between the plates
- The electric field between the plates
Answer: The capacitance
When a dielectric is inserted into an isolated (disconnected) capacitor, the charge remains constant. The capacitance increases by a factor K (C = KC₀), so the potential difference and electric field decrease (V = Q/C, E = V/d). Thus only the capacitance increases.
12. An electric dipole of dipole moment p is placed at the origin along the x-axis. The electric potential at a point (r, θ) far away from the dipole is given by V = (k p cosθ)/r². What is the potential at a point on the equatorial line of the dipole?
- k p / r²
- -k p / r²
- 0
- k p cosθ / r² (with θ = 90°)
Answer: 0
On the equatorial line, the angle θ between the dipole axis and the position vector is 90°. Since cos 90° = 0, V = (k p × 0)/r² = 0. Option D is just the general formula, but at θ = 90° it equals zero. Options A and B are the axial line potentials (θ = 0° or 180°).