Electric Charges and Fields Solutions Krishna home tuition

1. Electric Charges and Fields Solutions Introduction Electricity is an essential part of our daily lives, from powering our homes to running electronic devices. Understanding electric charges and fields is fundamental to mastering physics concepts, particularly in Class 12. This topic plays a crucial role in understanding electrostatics, which forms the foundation for various applications in electrical engineering and electronics. In this blog, we will discuss electric charges, Coulomb’s law, electric fields, and their solutions to help students grasp the concepts effectively. What are Electric Charges? Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. Charges are of two types: ● Positive Charge: Possessed by protons. ● Negative Charge: Possessed by electrons. Properties of Electric Charges 1. Like charges repel, and unlike charges attract. 2. Charge is conserved – it cannot be created or destroyed but can be transferred. 3. Charge is quantized – the smallest unit of charge is e = 1.6 × 10⁻¹⁹ C. 4. Charge is additive – the total charge is the algebraic sum of individual charges. 5. Charge is transferable – it can move from one object to another, especially in conductors. Coulomb’s Law Coulomb’s law defines the force between two point charges: F=kq1q2r2F = k \frac{q_1 q_2}{r^2} Where:

2. ● F = Force between charges (Newton) ● k = Coulomb’s constant (9 × 10⁹ Nm²/C²) ● q₁, q₂ = Magnitudes of charges (Coulombs) ● r = Distance between the charges (meters) Electric Field (E) An electric field is a region around a charged object where another charge experiences a force. It is defined as: E=FqE = \frac{F}{q} Where: ● E = Electric field (N/C) ● F = Force on test charge (N) ● q = Test charge (C) For a point charge: E=kQr2E = k \frac{Q}{r^2} Where Q is the charge creating the field. Solution Example: Problem: Find the electric field at a distance of 20 cm from a charge of 5μC. Solution: ● Q = 5 × 10⁻⁶ C ● r = 0.2 m Using the formula: E=(9×109)×(5×10−6)(0.2)2E = \frac{(9 × 10^9) × (5 × 10^{-6})}{(0.2)^2} E=(9×109)×(5×10−6)/0.04E = (9 × 10^9) × (5 × 10^{-6}) / 0.04 E=1.125×106 N/CE = 1.125 × 10^6 \, N/C So, the electric field is 1.125 × 10⁶ N/C. Electric Field Lines Electric field lines help visualize the electric field around a charge:

3. 1. Field lines originate from positive charges and terminate at negative charges. 2. They never intersect. 3. The density of lines represents field strength. 4. They are perpendicular to the surface of conductors. Electric Dipole An electric dipole consists of two equal and opposite charges separated by a distance d. Dipole Moment (p): p=qdp = qd The electric field at an axial position of a dipole: E=14πε0×2pr3E = \frac{1}{4\pi\varepsilon_0} \times \frac{2p}{r^3} The electric field at an equatorial position: E=14πε0×pr3E = \frac{1}{4\pi\varepsilon_0} \times \frac{p}{r^3} Gauss’s Law Gauss’s law states: ∮E⋅dA=qε0\oint E \cdot dA = \frac{q}{\varepsilon_0} Where: ● Φ = Electric flux ● E = Electric field ● A = Surface area ● q = Charge enclosed ● ε₀ = Permittivity of free space (8.85 × 10⁻¹² C²/Nm²) Conclusion Understanding electric charges and fields is crucial for mastering electrostatics. Concepts like Coulomb’s law, electric field, and Gauss’s law provide a strong foundation for advanced physics topics. Practicing numerical problems and conceptual questions is key to excelling in this subject. At Krishna Home Tuition, we provide expert home tuition in Mohali and Chandigarh to help students master these concepts with ease. Our tutors focus on personalized learning, problem-solving techniques, and conceptual clarity to boost student confidence and performance. ? Contact us today for one-on-one tutoring and make physics easy and enjoyable!