Electric Charges and Fields: CBSE Class 12 Physics Chapter 1 Guide

Fundamental Properties of Electric Charge

Electric Charge

An intrinsic property of elementary particles of matter which gives rise to electric force between various objects. It is a scalar quantity, with the SI unit being the Coulomb (C). There are two types: positive and negative.

Quantization of Charge

This principle states that the total charge on any body is always an integral multiple of a basic unit of charge, denoted by 'e' (the charge of an electron or proton). Mathematically, Q = ne, where n is an integer and e ≈ 1.602 × 10⁻¹⁹ C. Charge cannot exist in fractional multiples of 'e'.

Conservation of Charge

For an isolated system, the total electric charge remains constant. Charges can be created or destroyed, but only in equal and opposite pairs. For example, in pair production, a gamma-ray photon converts into an electron-positron pair, keeping the net charge zero.

Additivity of Charge

The total electric charge of a system is the algebraic sum of all the individual charges present in the system. While summing, the signs of the charges must be taken into account. For a system with charges q₁, q₂, q₃, ..., qₙ, the total charge is Q = q₁ + q₂ + q₃ + ... + qₙ.

Coulomb's Law and the Concept of Electric Field

How do charges interact? The answer lies in Coulomb's Law. Formulated by Charles-Augustin de Coulomb in the 1780s, this law quantifies the force between two stationary point charges. It states that the electrostatic force (F) between two charges (q₁ and q₂) is directly proportional to the product of the magnitude of the charges and inversely proportional to the square of the distance (r) between them.

Mathematically, in scalar form: F = k |q₁q₂| / r², where k is the electrostatic constant (k ≈ 9 × 10⁹ N m²/C²). This inverse square relationship is profound; it means the force weakens rapidly as the distance increases. The direction of the force is along the line joining the two charges. It's repulsive for like charges (+,+ or -,-) and attractive for unlike charges (+,-).

The concept of an Electric Field (E) was introduced to explain this 'action at a distance'. Instead of one charge directly acting on another, we say a charge creates an electric field in the space around it. Any other charge placed in this field experiences a force. The electric field at a point is defined as the force experienced by a unit positive test charge (q₀) placed at that point: E = F / q₀. Its SI unit is Newtons per Coulomb (N/C). For a point charge q, the electric field at a distance r is E = k|q|/r². The field provides a map of the force a charge would feel at any point in space.

Worked Examples on Electric Force and Field

• Example 1: Force between two charges Two point charges, q₁ = +2 μC and q₂ = -4 μC, are separated by a distance of 30 cm in a vacuum. Calculate the magnitude and nature of the force between them. Step 1: Identify the given values and convert to SI units. q₁ = +2 μC = 2 × 10⁻⁶ C q₂ = -4 μC = -4 × 10⁻⁶ C r = 30 cm = 0.3 m k = 9 × 10⁹ N m²/C² Step 2: Apply Coulomb's Law formula. F = k |q₁q₂| / r² F = (9 × 10⁹) × |(2 × 10⁻⁶) × (-4 × 10⁻⁶)| / (0.3)² Step 3: Calculate the magnitude. F = (9 × 10⁹) × (8 × 10⁻¹²) / 0.09 F = (72 × 10⁻³) / (9 × 10⁻²) F = 8 × 10⁻¹ = 0.8 N Step 4: Determine the nature of the force. Since the charges are of opposite signs (one positive, one negative), the force is attractive. Final Answer: The force between the charges is 0.8 N and it is attractive in nature.
• Example 2: Superposition Principle Three charges q₁ = +1 μC, q₂ = -2 μC, and q₃ = +3 μC are placed at the vertices of an equilateral triangle of side 10 cm. Find the net force on charge q₁. Step 1: Visualize the setup and draw a diagram. Place q₁ at the top vertex, q₂ at the bottom left, and q₃ at the bottom right. The angle at each vertex is 60°. Step 2: Calculate the force on q₁ due to q₂ (F₁₂). r = 10 cm = 0.1 m F₁₂ = k |q₁q₂| / r² = (9 × 10⁹) × |(1×10⁻⁶)(-2×10⁻⁶)| / (0.1)² = 1.8 N. Since q₁ is positive and q₂ is negative, this force is attractive, directed from q₁ towards q₂. Step 3: Calculate the force on q₁ due to q₃ (F₁₃). F₁₃ = k |q₁q₃| / r² = (9 × 10⁹) × |(1×10⁻⁶)(3×10⁻⁶)| / (0.1)² = 2.7 N. Since both q₁ and q₃ are positive, this force is repulsive, directed away from q₃ along the line joining them. Step 4: Find the vector sum of the forces. The angle between the vectors F₁₂ and F₁₃ is 120° (60° inside the triangle + 60° away from q₃). Use the law of cosines for vector addition: F_net = √(F₁₂² + F₁₃² + 2F₁₂F₁₃cosθ) F_net = √(1.8² + 2.7² + 2(1.8)(2.7)cos(120°)) cos(120°) = -0.5 F_net = √(3.24 + 7.29 + 2(1.8)(2.7)(-0.5)) F_net = √(10.53 - 4.86) = √5.67 ≈ 2.38 N Final Answer: The net force on charge q₁ is approximately 2.38 N. The direction can be found using the law of sines.

Exam Traps and Important Pointers

Scoring well in this chapter requires careful attention to detail. Here are some common traps students fall into:

• Vector Nature of Force & Field: Don't just add magnitudes of forces or fields algebraically. Always treat them as vectors. Use the principle of superposition and vector addition (parallelogram law or component method) when dealing with multiple charges.
• Sign Conventions: The signs of charges are crucial. In Coulomb's law, the signs determine if the force is attractive or repulsive. When calculating total charge or electric potential (in the next chapter), you must use the signs.
• Units: Always convert all quantities to SI units before calculation (e.g., cm to m, μC to C). A mistake here can throw off the entire answer.
• Gauss's Law Application: Remember that Gauss's Law is most useful for charge distributions with high symmetry (spheres, infinite lines, infinite sheets). Choosing the correct Gaussian surface that matches the symmetry is the key step. For non-symmetric distributions, you must revert to integrating the electric field from first principles.

Practice Questions with Solutions

• Q: How many electrons must be removed from a conductor so that it acquires a positive charge of 3.2 nC? A: Step 1: Identify the given total charge (Q) and the elementary charge (e). Q = 3.2 nC = 3.2 × 10⁻⁹ C. e = 1.6 × 10⁻¹⁹ C. Step 2: Use the formula for quantization of charge, Q = ne, where n is the number of electrons. Step 3: Rearrange the formula to solve for n: n = Q / e. n = (3.2 × 10⁻⁹ C) / (1.6 × 10⁻¹⁹ C) = 2 × 10¹⁰. Final answer: 2 × 10¹⁰ electrons must be removed.
• Q: Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them is the electric field zero? A: Step 1: Let the point P be at a distance 'x' from the +4q charge. The distance from the +q charge will be (30 - x) cm. At point P, the electric field due to +4q (E₁) must be equal and opposite to the electric field due to +q (E₂). Step 2: Set the magnitudes of the electric fields equal: E₁ = E₂. k(4q) / x² = k(q) / (30 - x)². Step 3: Solve for x. The 'kq' terms cancel out. 4 / x² = 1 / (30 - x)². Take the square root of both sides: 2 / x = 1 / (30 - x). 2(30 - x) = x => 60 - 2x = x => 3x = 60 => x = 20 cm. Final answer: The electric field is zero at a distance of 20 cm from the +4q charge and 10 cm from the +q charge.
• Q: An electric dipole with dipole moment 4 × 10⁻⁹ C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 10⁴ N/C. Calculate the magnitude of the torque acting on the dipole. A: Step 1: Identify the given values. Dipole moment (p) = 4 × 10⁻⁹ C m. Electric field (E) = 5 × 10⁴ N/C. Angle (θ) = 30°. Step 2: Recall the formula for torque (τ) on a dipole in a uniform electric field: τ = pE sinθ. Step 3: Substitute the values into the formula. τ = (4 × 10⁻⁹ C m) × (5 × 10⁴ N/C) × sin(30°). Step 4: Calculate the result. sin(30°) = 0.5. τ = (20 × 10⁻⁵) × 0.5 = 10 × 10⁻⁵ = 1 × 10⁻⁴ N m. Final answer: The magnitude of the torque is 1 × 10⁻⁴ N m.
• Q: A spherical conductor of radius 12 cm has a charge of 1.6 × 10⁻⁷ C distributed uniformly on its surface. What is the electric field at a point 18 cm from the center of the sphere? A: Step 1: Identify the given values. Radius of sphere (R) = 12 cm = 0.12 m. Charge (Q) = 1.6 × 10⁻⁷ C. Distance of point (r) = 18 cm = 0.18 m. Step 2: Recognize that for a point outside a uniformly charged spherical shell, the sphere behaves as if all its charge is concentrated at the center. We can use the formula for the electric field of a point charge: E = kQ / r². Step 3: Substitute the values into the formula. E = (9 × 10⁹ N m²/C²) × (1.6 × 10⁻⁷ C) / (0.18 m)². E = (14.4 × 10²) / 0.0324 = 1440 / 0.0324. Step 4: Calculate the final value. E = 44444.4... N/C ≈ 4.44 × 10⁴ N/C. The direction is radially outward as the charge is positive. Final answer: The electric field at that point is approximately 4.44 × 10⁴ N/C.

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