1.The specific charge ($e/m$ ratio) of an electron is approximately ____________ $\text{C/kg}$.
Ans: $1.76 \times 10^{11}$
2.In a neutral, isolated atom, the number of protons is strictly ____________ to the number of electrons.
Ans: equal
3.A body has a charge of $-80 \mu\text{C}$. The number of excess electrons on it is:
Ans: (B) $5 \times 10^{14}$ [Hint: $n = q/e = (80 \times 10^{-6}) / (1.6 \times 10^{-19})$]
4.Which of the following charges cannot exist in nature?
Ans: (D) $5.4 \times 10^{-19} \text{ C}$ [Hint: It is not an integral multiple of $1.6 \times 10^{-19} \text{ C}$]
5.Calculate the total charge on an alpha ($\alpha$) particle.
Ans: An alpha particle is a helium nucleus ($2\text{ protons}$). Charge $q = +2e = 2 \times 1.6 \times 10^{-19} = +3.2 \times 10^{-19} \text{ C}$.
6.How many electrons must be removed from a neutral piece of metal to give it a positive charge of $1.0 \times 10^{-7} \text{ C}$?
Ans: $n = q/e = (1.0 \times 10^{-7}) / (1.6 \times 10^{-19}) = 6.25 \times 10^{11} \text{ electrons}$.
7.Does the mass of a body strictly change when it is given a negative charge? Explain briefly.
Ans: Yes. Giving a negative charge implies adding electrons. Since electrons have a finite mass ($9.1 \times 10^{-31}\text{ kg}$), the mass of the body increases very slightly.
8.Is it possible for two similarly charged bodies to attract each other? Give a reason.
Ans: Yes. If one charged body is significantly larger or has a much higher charge magnitude than the other, it can induce an opposite charge on the nearer face of the smaller body, leading to net attraction.
9.What is the macroscopic physical cause behind the quantization of electric charge?
Ans: The transfer of electrons during charging. Since only whole, individual electrons can be transferred from one body to another, charge is quantized.
10.Match the basic properties of charge with their defining expressions:
Ans: (a) $\rightarrow$ (iii) $q = \pm ne$
(b) $\rightarrow$ (i) $Q_{total} = q_1 + q_2 + ... + q_n$
(c) $\rightarrow$ (ii) $\Delta Q_{isolated\_system} = 0$
11.The value of the dielectric constant ($K$ or $\epsilon_r$) for pure water is approximately ____________.
Ans: $80$ (or exactly $81$ depending on temperature)
12.If the distance between two point charges is halved, the electrostatic force between them becomes ____________ times the original force.
Ans: $4$ (Four) times [$F \propto 1/r^2$]
13.The force between two charges in vacuum is $F$. If a thick brass plate is introduced completely between them, the force becomes:
Ans: (D) Zero [Hint: For metals, $K = \infty$, so $F_{medium} = F/K = F/\infty = 0$].
14.The ratio of electric force ($F_e$) to gravitational force ($F_g$) between an electron and a proton separated by a distance is of the order of:
Ans: (A) $10^{39}$
15.Two point charges of $+5 \mu\text{C}$ and $+10 \mu\text{C}$ are placed $20 \text{ cm}$ apart in a vacuum. Calculate the electrostatic force between them.
Ans: $F = k \frac{q_1 q_2}{r^2} = 9 \times 10^9 \frac{(5 \times 10^{-6})(10 \times 10^{-6})}{(0.2)^2} = \frac{450 \times 10^{-3}}{0.04} = 11.25 \text{ N}$ (Repulsive).
16.The electrostatic force on a small sphere of charge $0.4 \mu\text{C}$ due to another small sphere of charge $-0.8 \mu\text{C}$ in air is $0.2 \text{ N}$. What is the distance between the two spheres?
Ans: $r = \sqrt{k \frac{|q_1 q_2|}{F}} = \sqrt{9 \times 10^9 \frac{0.4 \times 10^{-6} \times 0.8 \times 10^{-6}}{0.2}} = \sqrt{144 \times 10^{-4}} = 12 \times 10^{-2} \text{ m} = 12 \text{ cm}$.
17.In the previous question, what is the magnitude of the force exerted on the second sphere by the first sphere?
Ans: $0.2 \text{ N}$. By Newton's third law, electrostatic forces act as action-reaction pairs.
18.Define $1 \text{ Coulomb}$ of charge strictly using Coulomb's law.
Ans: One Coulomb is that amount of charge which, when placed in a vacuum at a distance of $1\text{ meter}$ from an identical charge, repels it with a force of $9 \times 10^9 \text{ Newtons}$.
19.State two major limitations of Coulomb's law in electrostatics.
Ans: 1. It is strictly valid only for point charges. 2. It is applicable only when the charges are at rest (electrostatics).
20.Match the medium to its corresponding approximate dielectric constant ($K$):
Ans: (a) $\rightarrow$ (iii) $1$ (approx)
(b) $\rightarrow$ (i) $80$
(c) $\rightarrow$ (ii) Infinity ($\infty$)
21.The electric field inside a solid conducting sphere given a charge $Q$ is mathematically ____________.
Ans: Zero
22.A uniformly spaced, parallel set of electric field lines represents a ____________ electric field.
Ans: uniform
23.The dimensional formula of Electric Field Intensity is:
Ans: (B) $[M L T^{-3} A^{-1}]$ [Hint: $E = F/q = [M L T^{-2}] / [A T]$]
24.A proton and an electron are placed in the exact same uniform electric field. They will experience:
Ans: (B) Equal forces in magnitude but opposite in direction. (Accelerations will differ because their masses differ).
25.Calculate the magnitude of the electric field at a distance of $5 \text{ cm}$ from a point charge of $10 \mu\text{C}$.
Ans: $E = k \frac{q}{r^2} = 9 \times 10^9 \frac{10 \times 10^{-6}}{(0.05)^2} = \frac{90 \times 10^3}{0.0025} = 3.6 \times 10^7 \text{ N/C}$.
26.Determine the electric field strength required to just balance the weight of an electron.
Ans: Electrostatic force must balance weight: $eE = mg$.
$E = \frac{mg}{e} = \frac{9.1 \times 10^{-31} \times 9.8}{1.6 \times 10^{-19}} \approx 5.57 \times 10^{-11} \text{ V/m}$ (Directed downwards to push electron up).
27.Calculate the force experienced by an alpha particle placed in an electric field of $3 \times 10^4 \text{ N/C}$.
Ans: $F = qE = (2 \times 1.6 \times 10^{-19}) \times (3 \times 10^4) = 9.6 \times 10^{-15} \text{ N}$.
28.Why do electric field lines never form closed continuous loops?
Ans: Because the electrostatic field is a conservative field, implying the net work done moving a charge along a closed loop must be zero, which prohibits closed field lines.
29.Can a charged particle always move along the path of an electric field line? Explain.
Ans: No. It will only follow the field line perfectly if the field line is perfectly straight (and starts from rest). For curved lines, inertia will cause it to deviate tangentially.
30.Match the charge configuration with its corresponding electric field line pattern:
Ans: (a) $\rightarrow$ (iii) Radial, pointing outwards
(b) $\rightarrow$ (i) Radial, pointing inwards
(c) $\rightarrow$ (ii) Parallel, equally spaced straight lines
31.The net translational force acting on an electric dipole placed in a *uniform* electric field is always ____________.
Ans: Zero
32.Electric flux is a ____________ physical quantity (scalar/vector).
Ans: scalar
33.A practical non-SI unit often used for electric dipole moment is the:
Ans: (B) Debye ($1 \text{ Debye} \approx 3.33 \times 10^{-30} \text{ C}\cdot\text{m}$)
34.If an electric dipole is placed in a non-uniform electric field, it generally experiences:
Ans: (C) Both a net force and a torque
35.Two point charges of $\pm 10 \mu\text{C}$ are placed $5 \text{ mm}$ apart. Calculate the magnitude of the electric dipole moment.
Ans: $p = q \times (2a) = (10 \times 10^{-6}) \times (5 \times 10^{-3}) = 5 \times 10^{-8} \text{ C}\cdot\text{m}$.
36.Using the dipole from Q35, calculate the electric field at a point on its axial line at a distance of $15 \text{ cm}$ from its center.
Ans: For a short dipole ($r \gg 2a$): $E = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3} = (9 \times 10^9) \frac{2 \times (5 \times 10^{-8})}{(0.15)^3} = \frac{900}{0.003375} \approx 2.67 \times 10^5 \text{ N/C}$.
37.A circular plane sheet of radius $10 \text{ cm}$ is placed in a uniform electric field of $5 \times 10^5 \text{ N/C}$, making an angle of $60^\circ$ with the field lines. Calculate flux.
Ans: Angle with field is $60^\circ \implies$ Angle with normal ($\theta$) is $90^\circ - 60^\circ = 30^\circ$.
Area $A = \pi r^2 = \pi (0.1)^2 = 0.01\pi \text{ m}^2$.
$\Phi = EA \cos\theta = (5 \times 10^5)(0.01\pi)(\cos 30^\circ) = 5000\pi (\sqrt{3}/2) \approx 1.36 \times 10^4 \text{ N m}^2/\text{C}$.
38.Define an "ideal" or "point" electric dipole.
Ans: An ideal dipole is one where the charges $q \to \infty$ and distance $2a \to 0$, such that the dipole moment $p = q \times 2a$ remains a finite constant. It has practically zero size.
39.What is the work done in rotating a dipole from its position of stable equilibrium to unstable equilibrium in a uniform electric field?
Ans: Stable ($\theta_1 = 0^\circ$), Unstable ($\theta_2 = 180^\circ$).
$W = pE(\cos\theta_1 - \cos\theta_2) = pE(1 - (-1)) = 2pE$.
40.Match the angle $\theta$ (between $\vec{p}$ and $\vec{E}$) to the resulting torque magnitude ($\tau$):
Ans: (a) $\rightarrow$ (iii) Zero torque (Stable equilibrium)
(b) $\rightarrow$ (i) Maximum torque ($pE$)
(c) $\rightarrow$ (ii) Zero torque (Unstable equilibrium)
41.The physical SI unit of surface charge density ($\sigma$) is ____________.
Ans: $\text{C/m}^2$ (Coulombs per square meter)
42.A Gaussian surface must ideally be chosen such that the electric field is either ____________ or ____________ to the surface at every point.
Ans: parallel, perpendicular
43.If a charge $q$ is placed exactly at the center of one face of a cube, the total flux passing through the cube is:
Ans: (B) $q/2\epsilon_0$ [Hint: The charge is shared equally by two cubes placed face-to-face].
44.Gauss's law in electrostatics is fundamentally a consequence of:
Ans: (C) The inverse square dependence on distance (in Coulomb's law).
45.A spherical Gaussian surface encloses a net charge of $8.854 \times 10^{-8} \text{ C}$. Calculate the total electric flux passing through the surface.
Ans: $\Phi = \frac{q_{in}}{\epsilon_0} = \frac{8.854 \times 10^{-8}}{8.854 \times 10^{-12}} = 10^4 \text{ N m}^2/\text{C}$.
46.If the radius of the Gaussian surface in the previous question is doubled, how would the electric flux change?
Ans: The flux remains the same ($10^4 \text{ N m}^2/\text{C}$) because flux depends only on the enclosed charge, not on the radius/size of the Gaussian surface.
47.A point charge $+10 \mu\text{C}$ is at a distance $5 \text{ cm}$ directly above the center of a square of side $10 \text{ cm}$. What is the magnitude of the electric flux through the square?
Ans: The square can be treated as one face of a cube of side $10 \text{ cm}$ with the charge at its center. Flux through one face = $\frac{1}{6} \frac{q}{\epsilon_0} = \frac{1}{6} \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.88 \times 10^5 \text{ N m}^2/\text{C}$.
48.Is Gauss's law valid for a non-uniform electric field?
Ans: Yes, Gauss's law is universally valid for any closed surface and any electrostatic field (uniform or non-uniform), though it is mathematically useful mostly for symmetric fields.
49.What is the net electric flux through a closed surface if it completely encloses three identical electric dipoles?
Ans: Zero. The net charge of an electric dipole is zero. Thus, for three dipoles, $Q_{enclosed} = 3 \times 0 = 0 \implies \Phi = 0$.
50.Match the type of charge distribution to its appropriate standard symbol:
Ans: (a) $\rightarrow$ (ii) $\lambda$
(b) $\rightarrow$ (iii) $\sigma$
(c) $\rightarrow$ (i) $\rho$
51.The electric field just outside a charged macroscopic conducting plate is given by the formula ____________.
Ans: $\sigma/\epsilon_0$
52.The electric field at the surface of a charged spherical shell of radius $R$ and total charge $Q$ is ____________.
Ans: $\frac{1}{4\pi\epsilon_0} \frac{Q}{R^2}$
53.Electric field intensity due to a long straight uniformly charged wire varies with distance $r$ as:
Ans: (B) $1/r$
54.Two large, thin metal plates are parallel and close to each other. On their inner faces... The electric field between the plates is:
Ans: (C) $\sigma/\epsilon_0$ [Hint: Fields from both plates add up: $\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}$]
55.An infinite line charge produces a field of $9 \times 10^4 \text{ N/C}$ at a distance of $2 \text{ cm}$. Calculate the linear charge density.
Ans: $E = \frac{\lambda}{2\pi\epsilon_0 r} \implies \lambda = E \times 2\pi\epsilon_0 r = \frac{E \times r}{2 \times (1/4\pi\epsilon_0)} = \frac{(9 \times 10^4) \times (0.02)}{2 \times 9 \times 10^9} = 10^{-7} \text{ C/m}$ (or $0.1 \mu\text{C/m}$).
56.Referring to Q54, what is the electric field in the outer regions (above the top plate and below the bottom plate)?
Ans: Zero. The fields produced by the positive plate and the negative plate are equal in magnitude but opposite in direction in the outer regions, hence cancelling each other out.
57.Why is the electric field strictly zero everywhere inside a hollow charged conductor under electrostatic conditions?
Ans: Charges reside entirely on the outer surface of a conductor to minimize repulsion. Since $Q_{enclosed} = 0$ for any Gaussian surface inside the conductor, Gauss's law dictates that $E = 0$ inside.
58.Sketch a generic graph showing the variation of electric field $E$ with distance $r$ from the center of a uniformly charged thin spherical shell of radius $R$.
Ans: Graph properties: $E = 0$ from $r=0$ to $r=R$ (on the x-axis). At $r=R$, $E$ jumps to a maximum $E_{max} = \frac{kQ}{R^2}$. For $r > R$, the curve decays parabolically following $E \propto 1/r^2$.
59.Can we strictly use Gauss's law to analytically find the exact electric field of a *finite* line charge at any arbitrary point? Briefly explain.
Ans: No. For a finite line charge, the electric field is not symmetric (not purely radial everywhere), and edge effects exist. This makes choosing a simple Gaussian surface where $E$ is constant and parallel to the area vector impossible.
60.Match the geometrical charge source to how its electric field $E$ depends on distance $r$ (for outside points):
Ans: (a) $\rightarrow$ (ii) $E \propto 1/r^2$
(b) $\rightarrow$ (iii) $E \propto 1/r$
(c) $\rightarrow$ (i) $E \propto 1/r^3$
