Electrostatics
Each of the following questions comprises two statements. The assertion is expressed by Statement 1 and the reason is expressed by Statement 2.
Each question has four options (1), (2), (3), and (4) as given below, out of which only ONE is correct.
- Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
- Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
- Statement 1 is true and Statement 2 is false.
- Statement 1 is false and Statement 2 is true.
- Both Statement 1 and Statement 2 are false.
Some questions are preceded, by a supporting paragraph in addition to the statements.
Question 1. Statement 1 A charged particle free to move in an electric field always moves along the electric field lines.
Statement 2 The electric field lines diverge from a positive charge and converge to a negative charge.
Answer: 4. Statement 1 is false and Statement 2 is true.
A charged particle when released from rest in an electrostatic field always travels along the field. But when its initial velocity makes some angle 0 with the field direction, it follows a parabolic path. Electric field lines diverge from a positive charge.
Question 2. Statement 1 If a conductor is given charge then no excess inner charge appears;
Statement 2 The electric field inside a conductor is always zero.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
The charge given to a conductor totally remains on the outer surface and no part of it appears on the inner surface. The electric field inside a conductor is zero as a consequence of Gauss’s theorem and the absence of an inner charge.
Question 3. Statement 1 An electrostatic field line never forms a closed loop.
Statement 2 An electrostatic field is a conservative force field.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
Electrostatic field lines do not form a closed loop. In a dipole, they start at the positive charge and terminate at the negative charge. This corresponds to the conservative nature of an electrostatic field for which the work done for a closed path is zero.
Question 4. Statement 1 Gauss’s law cannot be used to calculate the electric field near an electric dipole.
Statement 2 An electric dipole does not have symmetrical charge distribution.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
Gauss’s law is applicable when the charge distribution is cylindrical, spherical, or planar. It cannot be applied for an asymmetric charge distribution like that of an electric dipole.
Question 5. Statement 1 The Nuclear force is a dominating force in nature.
Statement 2 The Coulomb force is weaker than the gravitational force.
Answer: 3. Statement 1 is true and Statement 2 is false.
The nuclear forces are the strongest in nature, whereas the Coulomb force is much stronger than the gravitational force.
Question 6. Statement 1 On going away from a point charge or a small electric dipole, the electric field decreases at the same rate in both cases.
Statement 2 An electric field is inversely proportional to the square of the distance from the charge and the cube of the distance from the electric dipole.
Answer: 4. Statement 1 is false and Statement 2 is true.
The electric field due to a dipole \(\propto \frac{1}{r^3}\), and that due to a monopole \(\propto \frac{1}{r^2}\).
∴ \(\left(\frac{d E}{d r}\right)_{\text {dipole }} \propto \frac{1}{r^4} \text { and }\left(\frac{d E}{d r}\right)_{\text {monopole }} \propto \frac{1}{r^3}\)
Question 7. Statement 1 A Gaussian surface is considered carefully.
Statement 2 The point where the electric field is to be calculated should be within the surface.
Answer: 5. Both Statement 1 and Statement 2 are false.
A Gaussian surface is symmetrical about the given charge distribution, and the point where the field is to be found is outside the Gaussian surface.
Question 8. Statement 1 In the cavity within a conductor, the electric field is zero.
Statement 2 Charges with a conductor always reside only at its outer surface.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
The electric field in the cavity of a charged conductor is always zero because no charge exists on the inner surface. Both of these facts are consequences of Gauss’s theorem.
Question 9. Statement 1 Gauss’s law shows a diversion when the inverse square law is not obeyed.
Statement 2 Gauss’s law is a consequence of the conservation of charges.
Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
Gauss’s law is a consequence of the inverse-square law. Charge cannot be created nor be destroyed, it always remains conserved.
Question 10. Statement 1 The acceleration of a charged particle in a uniform electric field does not depend on the velocity of the charged particle.
Statement 2 Charge is an invariant quantity, which means that the amount of charge with a particle does not depend on the frame of reference.
Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
Force = F = qE; acceleration = \(a=\frac{F}{m}=\frac{q E}{m}\)
Since q, E, and m are constants, the acceleration a is also a constant and independent of the velocity of the charged particle. Charge is always invariant.
Question 11. Statement 1 If an electron and a proton possessing equal KEs enter an electric field in a particular direction, the path of the electron will be more curved than that of the proton.
Statement 2 An electron describes a larger curve due to its small mass.
Answer: 5. Both Statement 1 and Statement 2 are false.
When a charged particle is projected at some angle θ (≠ 0° or π), it follows a parabolic path, for which the curvature varies from point to point.
At a given point, \(F=q E=\frac{m v^2}{r}=\frac{2 \mathrm{KE}}{r}\)
An electron and a proton have the same KE and the same charge, so the radius of the curvature r will be the same and the path followed will be the same.
Question 12. Statement 1 Electrons move away from a region of lower potential to a region of higher potential.
Statement 2 An electron has a negative charge.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
Electrons are negatively charged particles that move against the direction of the electric field. An electric field is directed from a higher potential to a lower potential.
Question 13. Statement 1 The electrical potential of the earth is taken as zero.
Statement 2 No electric field exists on the earth’s surface.
Answer: 3. Statement 1 is true and Statement 2 is false.
The earth is a very large conducting body where the addition or subtraction of charges is insignificant. Hence, its potential is taken to be zero. Near the surface of the earth, the magnitude of the electric field is around 100 V m-1.
Question 14. Statement 1 Electric field lines are perpendicular to a conducting surface.
Statement 2 Electric field lines are perpendicular to an equipotential surface.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
A conducting surface is equipotential, on which the electrical field is directed perpendicular to its surface.
Question 15. Statement 1 Two adjacent spherical conductors carrying the same amount of positive charge have a potential difference between them.
Statement 2 The potential to which a conductor is raised depends on the charge.
Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
The potential of a spherical conductor of radius R is \(V=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{R}\).
If the charge is the same and the radii of the spheres are unequal then their potentials will be different. The potential V depends on both the charge Q and the radius R.
Question 16. Statement 1 An electric field is preferred to a magnetic field for deflecting the electron beam in a television picture tube.
Statement 2 An electric field requires a low voltage.
Answer: 4. Statement 1 is false and Statement 2 is true.
In television picture tubes, a magnetic field is preferred to an electric field for deflecting the electron beam.
The electric field \(E=\frac{V}{d}\) requires a high voltage to produce the required field.
Question 17. Statement 1 If a dielectric is placed in an external field then the field inside the dielectric will be less than the field outside.
Statement 2 The electric field will induce a dipole moment opposite to the field direction.
Answer: 3. Statement 1 is true and Statement 2 is false.
Dielectric polarization produces a field in a direction opposite to the externally applied field.
Polarization produces dipoles whose moments (-ve to +ve) are directed along the externally applied field.
Question 18. Statement 1 The electrostatic potential energy of a spherical shell with a uniformly distributed negative charge is positive.
Statement 2 Two similar charges repel each other.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
Potential energy = work done
∴ \(\int d W=\int V d q=\int_0^{-Q_1} \frac{1}{4 \pi \varepsilon_0 R}(-q)(-d q)=\frac{Q^2}{8 \pi \varepsilon_0 R}=\frac{( \pm Q)^2}{8 \pi \varepsilon_0 R}\)
This is positive for both positive and negative charges. Potential energy is positive for a repulsive field (repulsive force between two like charges).
Question 19. Statement 1 When an uncharged parallel-plate capacitor is charged by connecting it to a cell, the heat produced in the circuit is equal to the energy stored in the capacitor.
Statement 2 The charge on a parallel-plate capacitor means equal and opposite charges on its inner faces.
Answer: 2. Statement 1 is true; Statement 2 is true; but Statement 2 is not a correct explanation of Statement 1.
Total energy delivered by the cell = Qε .
Energy stored in the capacitor = \(\frac{Q^2}{2 C}=\frac{Q \varepsilon}{2}\).
Heat developed = \(\mathrm{Q} \varepsilon-\frac{\mathrm{Q} \varepsilon}{2}=\frac{\mathrm{Q} \varepsilon}{2}\).
A charged capacitor has equal and opposite charges, which produce an electrostatic field between the plates.
Question 20. Statement 1 It is possible for a charged particle to move in a circular path around a uniformly charged, long, straight conductor.
Statement 2 The electrostatic force on a moving charged particle is directed toward the conductor.
Answer: 1. Statement 1 is true; Statement 2 is true; and Statement 2 is a correct explanation of Statement 1.
The electrostatic field due to a long, straight-charged conductor is radial, which can provide a centripetal force to the charged particle moving along the circle/with the axis as the conductor.
