Chapter 1 - Electric Charges and Fields

Time: 2 Hours | Max Marks: 50

Section A: Multiple Choice Questions (10 x 1 Mark)

  1. If a glass rod is rubbed with silk, the glass rod acquires a:

    a) Negative charge

    b) Positive charge

    c) No charge

    d) Depends on the length of the rod

  2. The SI unit of electric flux is:

    a) N/CN/C

    b) V⋅mV \cdot m

    c) N⋅m2/CN \cdot m^2/C

    d) Both b and c

  3. When a point charge qq is placed at the corner of a cube, the total flux through the cube is:

    a) qϵ0\frac{q}{\epsilon_0}

    b) q2ϵ0\frac{q}{2\epsilon_0}

    c) q8ϵ0\frac{q}{8\epsilon_0}

    d) Zero

  4. The quantization of charge means that:

    a) Charge cannot be destroyed

    b) Charge exists in discrete packets of ee

    c) Charge is always positive

    d) Total charge of a system is conserved

  5. The electric field inside a perfectly conducting hollow sphere is:

    a) Maximum

    b) Minimum

    c) Zero

    d) Dependent on the radius

  6. Which of the following is not a property of electric field lines?

    a) They form closed loops

    b) They start from positive and end at negative

    c) They never intersect

    d) They are continuous curves

  7. If the distance between two charges is doubled, the electrostatic force becomes:

    a) Double

    b) Half

    c) Four times

    d) One-fourth

  8. The direction of an electric dipole moment is:

    a) From positive to negative

    b) From negative to positive

    c) Perpendicular to the axis

    d) Circular

  9. A surface that has the same potential at every point is called:

    a) Isotropic surface

    b) Equipotential surface

    c) Gaussian surface

    d) Linear surface

  10. The permittivity of free space (ϵ0\epsilon_0) has the value:

    a) 9×109 N m2/C29 \times 10^9 \text{ N m}^2/\text{C}^2

    b) 1.6×10−19 C1.6 \times 10^{-19} \text{ C}

    c) 8.854×10−12 C2/(N m2)8.854 \times 10^{-12} \text{ C}^2/(\text{N m}^2)

    d) 11

Section B: Short Answer Questions (8 x 2 Marks)

  1. Define Coulomb’s Law in vector form.

  2. State two basic properties of electric charge.

  3. Why do two electric field lines never cross each other?

  4. Define "Electric Flux." Is it a scalar or vector quantity?

  5. What is an electric dipole? Give one example of a polar molecule.

  6. Briefly explain when the torque on a dipole in a uniform electric field is maximum.

  7. State Gauss’s Law for electrostatics.

  8. Why are electric field lines perpendicular to the surface of a conductor?

Section C: Long Answer Questions (3 x 5 Marks)

  1. Field of a Dipole: Derive an expression for the electric field intensity at a point on the axial line of an electric dipole.

  2. Continuous Charge Distribution: Using Gauss’s Law, derive the expression for the electric field due to an infinitely long straight uniformly charged wire. 3. Flux and Symmetry: Explain the concept of symmetry in finding flux. If a charge is placed at the corner of a cube, explain why the flux through the faces touching the charge is zero.

Section D: Numericals (5 x 3 Marks)

  1. Force Calculation: Two point charges q1=+4μCq_1 = +4 \mu\text{C} and q2=−6μCq_2 = -6 \mu\text{C} are placed 20 cm20\text{ cm} apart in a vacuum. Calculate the magnitude of the force between them.

  2. Dipole Torque: An electric dipole with a dipole moment of 5×10−9 C m5 \times 10^{-9}\text{ C m} is aligned at 45∘45^\circ with the direction of a uniform electric field of magnitude 3×104 N/C3 \times 10^4\text{ N/C}. Calculate the magnitude of the torque acting on the dipole.

  3. Flux through a Sphere: A point charge of 17.7μC17.7 \mu\text{C} is at the center of a cubic Gaussian surface 15 cm15\text{ cm} on edge. What is the net electric flux through the surface? (Use ϵ0=8.85×10−12 C2/N m2\epsilon_0 = 8.85 \times 10^{-12} \text{ C}^2/\text{N m}^2).

  4. Field Strength: Calculate the electric field at a point 5 cm5\text{ cm} away from a point charge of 25μC25 \mu\text{C} in a vacuum.

  5. System of Charges: Four point charges qA=3μCq_A = 3 \mu\text{C}, qB=−7μCq_B = -7 \mu\text{C}, qC=3μCq_C = 3 \mu\text{C}, and qD=−7μCq_D = -7 \mu\text{C} are located at the corners of a square ABCD of side 10 cm10\text{ cm}. What is the force on a charge of 1μC1 \mu\text{C} placed at the center of the square?

Here are the solutions for the beginner-friendly practice paper on Electric Charges and Fields.


Section A: Multiple Choice Questions (Answers)

  1. b) Positive charge (Electrons move from the glass to the silk).

  2. d) Both b and c (N⋅m2/CN \cdot m^2/C is the standard unit; V⋅mV \cdot m is its equivalent).

  3. c) q8ϵ0\frac{q}{8\epsilon_0}.

  4. b) Charge exists in discrete packets of ee (Q=neQ = ne).

  5. c) Zero (Charges reside only on the outer surface).

  6. a) They form closed loops (This is a property of magnetic field lines, not electric).

  7. d) One-fourth (Inverse square law: F∝1/r2F \propto 1/r^2).

  8. b) From negative to positive.

  9. b) Equipotential surface.

  10. c) 8.854×10−12 C2/(N m2)8.854 \times 10^{-12} \text{ C}^2/(\text{N m}^2).


Section B: Short Answer Questions (Answers)

  1. Coulomb’s Law (Vector Form): F⃗12=14πϵ0q1q2r2r^12\vec{F}_{12} = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2} \hat{r}_{12}. It shows that the force acts along the line joining the two charges.

  2. Properties of Charge:

    • Additivity: Total charge is the algebraic sum of individual charges.

    • Conservation: Charge can neither be created nor destroyed in an isolated system.

  3. Field Line Crossing: If they crossed, there would be two different directions of the electric field at the point of intersection, which is physically impossible.

  4. Electric Flux: The total number of electric field lines passing through a given area. It is a scalar quantity.

  5. Electric Dipole: A pair of equal and opposite charges separated by a small distance. Example: H2OH_2O or HClHCl.

  6. Maximum Torque: Torque (τ=pEsin⁡θ\tau = pE \sin \theta) is maximum when the dipole is perpendicular (θ=90∘\theta = 90^\circ) to the field.

  7. Gauss’s Law: The total electric flux through any closed surface is equal to 1/ϵ01/\epsilon_0 times the net charge enclosed by that surface.

  8. Perpendicular Lines: If the field were not perpendicular, there would be a component of the field along the surface, causing charges to move (creating a current), which does not happen in electrostatics.


Section C: Long Answer Questions (Hints)

  1. Axial Line Derivation:

    • Place a point PP at distance rr from the center of the dipole (length 2a2a).

    • Enet=E+q−E−qE_{net} = E_{+q} - E_{-q}.

    • Result: E=14πϵ02pr(r2−a2)2E = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2 - a^2)^2}. For a short dipole (r≫ar \gg a): E=2p4πϵ0r3E = \frac{2p}{4\pi\epsilon_0 r^3}.

  2. Infinitely Long Wire:

    • Choose a cylindrical Gaussian surface of radius rr and length ll.

    • Φ=E(2πrl)\Phi = E(2\pi rl).

    • By Gauss’s Law, E(2πrl)=λlϵ0E(2\pi rl) = \frac{\lambda l}{\epsilon_0}.

    • Result: E=λ2πϵ0rE = \frac{\lambda}{2\pi\epsilon_0 r}.

  3. Flux and Symmetry:

    • A charge at a corner is shared by 8 identical cubes.

    • By symmetry, the total flux qϵ0\frac{q}{\epsilon_0} is divided equally among them, giving q8ϵ0\frac{q}{8\epsilon_0} per cube.

    • For the three faces touching the charge, the field lines are parallel to the surface (area vector is perpendicular), so the flux is zero.


Section D: Numericals (Step-by-Step)

  1. Force:

    F=k∣q1q2∣r2=(9×109)(4×10−6)(6×10−6)(0.2)2=5.4 NF = k \frac{|q_1 q_2|}{r^2} = (9 \times 10^9) \frac{(4 \times 10^{-6})(6 \times 10^{-6})}{(0.2)^2} = \mathbf{5.4 \text{ N}}

  2. Torque:

    τ=pEsin⁡θ=(5×10−9)(3×104)sin⁡45∘=1.5×10−4×0.707≈1.06×10−4 N m\tau = pE \sin \theta = (5 \times 10^{-9})(3 \times 10^4) \sin 45^\circ = 1.5 \times 10^{-4} \times 0.707 \approx \mathbf{1.06 \times 10^{-4} \text{ N m}}

  3. Flux through Cube:

    Φ=qenclosedϵ0=17.7×10−68.85×10−12=2×106 N m2/C\Phi = \frac{q_{enclosed}}{\epsilon_0} = \frac{17.7 \times 10^{-6}}{8.85 \times 10^{-12}} = \mathbf{2 \times 10^6 \text{ N m}^2/C}

  4. Field Strength:

    E=kqr2=(9×109)(25×10−6)(0.05)2=0.2250.0025=9×107 N/CE = \frac{kq}{r^2} = \frac{(9 \times 10^9)(25 \times 10^{-6})}{(0.05)^2} = \frac{0.225}{0.0025} = \mathbf{9 \times 10^7 \text{ N/C}}

  5. Center Charge:

    • The charges at opposite corners (AA and CC) are both +3μC+3 \mu\text{C}. They pull/push the center charge with equal force in opposite directions, canceling out.

    • The charges at BB and DD are both −7μC-7 \mu\text{C}. They also exert equal and opposite forces.

    • Net Force = 0 N.

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