Electric Charges and Fields Class 12 NCERT Solutions, PYQs & Notes
⚡ Introduction
Physics ki journey shuru hoti hai Electrostatics se, jahan hum stationary charges ke baare mein padhte hain. Is chapter mein hum samjhenge ki charges kya hote hain, Coulomb's law kaise kaam karta hai, Electric field lines ki properties kya hain, aur sabse important—Gauss's Law jo complex problems ko chutkiyon mein solve kar deta hai. Board exams ke liye derivations aur numericals dono is chapter se kaafi aate hain.
🔑 Key Formulas & Concepts
- Quantization of Charge: $q = \pm ne$, where $e = 1.6 \times 10^{-19}$ C.
- Coulomb's Law: The force between two point charges is given by: $$F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$$
- Electric Field ($E$): Force per unit positive test charge. $E = \frac{F}{q_0}$.
- Electric Dipole Moment ($p$): $p = q \times 2a$, directed from negative to positive charge.
- Gauss's Law: The total electric flux through a closed surface is $1/\epsilon_0$ times the total charge enclosed: $$\Phi = \oint \mathbf{E} \cdot d\mathbf{A} = \frac{q_{enclosed}}{\epsilon_0}$$
đź“š Part 1: Detailed NCERT Solutions
Q1: What is the force between two small charged spheres having charges of $2 \times 10^{-7}$ C and $3 \times 10^{-7}$ C placed 30 cm apart in air?
Ans:
Given: $q_1 = 2 \times 10^{-7}$ C, $q_2 = 3 \times 10^{-7}$ C, $r = 30 \text{ cm} = 0.3 \text{ m}$.
Using Coulomb’s Law: $F = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r^2}$
$F = (9 \times 10^9) \times \frac{(2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.3)^2}$
$F = \frac{9 \times 6 \times 10^{-5}}{0.09} = 6 \times 10^{-3} \text{ N}$ (Repulsive).
Q2: Define Electric Flux. Write its SI unit.
Ans: Electric Flux ($\Phi$) is the measure of the total number of electric field lines passing through a given area. It is defined as the dot product of the electric field vector and the area vector.
$$\Phi = \mathbf{E} \cdot \mathbf{A} = EA \cos \theta$$
SI Unit: N m²/C or Volt-meter (V m).
Q3: Why can two electric field lines never cross each other?
Ans: At any point on an electric field line, the tangent gives the direction of the electric field at that point. If two lines crossed, there would be two tangents at the point of intersection, implying two different directions for the electric field at the same point, which is physically impossible.
🔥 Part 2: 5 Most Repeated PYQs (Board Favorites)
Q1: Derive an expression for the electric field at a point on the axial line of an electric dipole.
Ans: Consider a dipole with charges $-q$ and $+q$ separated by distance $2a$. Let $P$ be a point on the axial line at distance $r$ from the center.
Field due to $+q$: $E_+ = \frac{1}{4\pi\epsilon_0} \frac{q}{(r-a)^2}$ (Away from dipole)
Field due to $-q$: $E_- = \frac{1}{4\pi\epsilon_0} \frac{q}{(r+a)^2}$ (Towards dipole)
Net Field $E = E_+ - E_-$
$$E = \frac{q}{4\pi\epsilon_0} \left[ \frac{1}{(r-a)^2} - \frac{1}{(r+a)^2} \right]$$
After solving, for $r \gg a$:
$$E_{axial} = \frac{1}{4\pi\epsilon_0} \frac{2p}{r^3}$$
The direction is along the dipole moment vector $\mathbf{p}$.
Q2: State Gauss's Law. Using this law, derive the expression for the electric field due to an infinitely long straight uniformly charged wire.
Ans:
Gauss’s Law: The total flux through a closed surface is $1/\epsilon_0$ times the net charge enclosed.
Derivation: Consider a wire with linear charge density $\lambda$. Take a cylindrical Gaussian surface of radius $r$ and length $l$.
Total flux $\Phi = \oint \mathbf{E} \cdot d\mathbf{A} = E \times (2\pi rl)$ (since field is radial).
Charge enclosed $q = \lambda l$.
By Gauss Law: $E(2\pi rl) = \frac{\lambda l}{\epsilon_0}$
$$E = \frac{\lambda}{2\pi\epsilon_0 r}$$
Q3: What happens to a dipole when placed in a uniform electric field? Derive the expression for torque.
Ans: In a uniform electric field, the net force on the dipole is zero ($F_{net} = qE - qE = 0$). However, the two forces act at different points, creating a torque.
Torque $\tau = \text{Force} \times \text{Perpendicular distance}$
$\tau = (qE) \times (2a \sin \theta)$
Since $p = q \times 2a$:
$$\tau = pE \sin \theta \text{ or } \boldsymbol{\tau} = \mathbf{p} \times \mathbf{E}$$
Q4: Derive the electric field due to a uniformly charged infinite plane sheet.
Ans: Consider a sheet with surface charge density $\sigma$. Take a cylindrical Gaussian surface piercing the sheet.
Total Flux $\Phi = EA + EA = 2EA$ (through two end caps).
Charge enclosed $q = \sigma A$.
By Gauss Law: $2EA = \frac{\sigma A}{\epsilon_0}$
$$E = \frac{\sigma}{2\epsilon_0}$$
Note: The field is independent of the distance $r$ from the sheet.
Q5: Define 'Electric Dipole Moment'. Is it a scalar or a vector?
Ans: The electric dipole moment ($\mathbf{p}$) is the product of the magnitude of one of the charges ($q$) and the distance between them ($2a$).
Nature: It is a vector quantity. Its direction is from the negative charge to the positive charge.
⚡ Part 3: 15 Extra Descriptive Practice Questions (CBT Style)
What is meant by 'Conservation of Charge'?
Ans: The principle of conservation of charge states that the total charge of an isolated system remains constant. Charge can neither be created nor destroyed; it can only be transferred from one body to another.
What is a 'Point Charge'?
Ans: If the sizes of charged bodies are very small compared to the distances between them, we treat them as point charges. All the charge is assumed to be concentrated at a single point in space.
How does the force between two charges change if a dielectric medium (water) is placed between them?
Ans: The force decreases. The force in a medium is $F_m = F_{vacuum} / K$, where $K$ is the dielectric constant. For water, $K \approx 80$, so the force becomes 1/80th of its original value.
Define 'Linear Charge Density'.
Ans: It is the charge per unit length of a conductor. $\lambda = \frac{\Delta q}{\Delta l}$. Its SI unit is C/m.
What is the physical significance of Electric Field?
Ans: It tells us the force that would be experienced by a unit positive charge placed at that point. It helps in understanding how electromagnetic energy and signals travel through space.
Sketch the electric field lines for a single positive charge and a single negative charge.
Ans: For a positive charge ($q > 0$), lines are directed radially outwards. For a negative charge ($q < 0$), lines are directed radially inwards.
What is the work done in moving a test charge over a closed path in an electric field?
Ans: The work done is zero because the electrostatic field is a conservative field. The work depends only on initial and final positions, not the path taken.
Under what condition is the torque on a dipole maximum?
Ans: Since $\tau = pE \sin \theta$, torque is maximum when $\sin \theta = 1$, i.e., $\theta = 90^\circ$. This happens when the dipole is perpendicular to the electric field.
What is the net flux through a closed surface containing an electric dipole?
Ans: The net flux is zero. According to Gauss Law, flux depends on the net charge enclosed. A dipole has charges $+q$ and $-q$, so net charge is zero.
Define 'Surface Charge Density'.
Ans: It is the charge per unit surface area. $\sigma = \frac{\Delta q}{\Delta A}$. Its SI unit is C/m².
Can a body have a charge of $0.8 \times 10^{-19}$ C?
Ans: No. Charge is quantized ($q = ne$). The smallest possible independent charge is $e = 1.6 \times 10^{-19}$ C. Since $0.8 \times 10^{-19}$ is less than $e$, it is not possible.
What is a 'Gaussian Surface'?
Ans: It is an imaginary closed surface in three-dimensional space through which the flux of a vector field is calculated. We usually choose a surface that matches the symmetry of the charge distribution.
What is the direction of the electric field at a point on the equatorial line of a dipole?
Ans: The direction is opposite to the direction of the dipole moment vector ($\mathbf{p}$).
Define 1 Coulomb of charge.
Ans: One Coulomb is that charge which, when placed at a distance of 1 meter from an equal and similar charge in vacuum, repels it with a force of $9 \times 10^9$ Newtons.
What is 'Electrostatic Shielding'?
Ans: It is the phenomenon of protecting a certain region of space from external electric fields by enclosing it in a hollow conductor (Faraday cage). The electric field inside a hollow conductor is always zero.
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