PYQ Masterclass: Boards & NEET

High-Yield Exam Tips & Trends

To ace your Board exams and NEET, keep these secrets in mind:

1. The Inverse Square Logic: Most numericals in NEET revolve around how $F$ or $E$ changes when distance $r$ is altered. Remember: $F \propto 1/r^2$ and $E \propto 1/r^2$ for point charges, but $E \propto 1/r^3$ for dipoles.
2. Vector Addition: Don't just add magnitudes! For Board exams, always draw the vector diagram (parallelogram or triangle) and show the resultant direction. This is where most students lose marks.
3. Gauss's Law Shortcuts: If a question asks for flux through a 'part' of a surface (like one face of a cube or a hemisphere), look for symmetry first before doing any integration.
4. Field Line Properties: Boards love 'Conceptual' questions. Be ready to explain why field lines are perpendicular to conductors or why they never intersect.
5. Dipole Equilibria: Differentiate clearly between stable ($\theta = 0^\circ$) and unstable ($\theta = 180^\circ$) equilibrium. NEET frequently tests the work done in rotating between these states.

Question 1: Define the SI unit of electric charge. Write its relation with the charge of an electron. [CBSE]

Solution:

1. Definition: The SI unit of electric charge is the Coulomb (C). One Coulomb is defined as the amount of charge that flows through a conductor in 1 second when a steady current of 1 Ampere is maintained.
2. Relation: The charge on one electron (elementary charge) is approximately $e = 1.6 \times 10^{-19}$ C.
3. Quantization: Any charge $Q$ is an integral multiple of $e$ ($Q = ne$).

Question 2: Two point charges $+q$ and $-q$ are placed at a distance $d$ apart. What is the net force on a third charge $Q$ placed at the midpoint between them? [CBSE]

Solution:

1. Identify Directions: The positive charge $+q$ repels $Q$ toward the $-q$ charge. The negative charge $-q$ attracts $Q$ toward itself.
2. Check Magnitudes: Both forces act in the same direction (toward the negative charge).
3. Calculate Force: $F_1 = k qQ / (d/2)^2$ and $F_2 = k qQ / (d/2)^2$.
4. Result: $F_{net} = F_1 + F_2 = 2 \times [k qQ / (d^2/4)] = 8k qQ / d^2$ toward the negative charge.

Question 3: A point charge $q$ is placed at the center of a spherical Gaussian surface. How is the electric flux through the surface affected if the radius of the sphere is doubled? [CBSE]

Solution:

1. Gauss's Law: $\Phi = q_{enclosed} / \epsilon_0$.
2. Analysis: The formula shows that flux depends only on the net charge enclosed, not on the dimensions or radius of the surface.
3. Answer: The electric flux remains unchanged.

Question 4: An electric dipole is kept in a non-uniform electric field. What does it experience? [NEET]

Solution:

1. Force: In a non-uniform field, the field strength at $+q$ and $-q$ is different. Thus, the forces $qE_1$ and $-qE_2$ do not cancel out. Net force $\neq 0$.
2. Torque: Since the forces act at different lines of action, the dipole experiences a torque.
3. Answer: It experiences both a net force and a torque.

Question 5: Plot a graph showing the variation of electric field $E$ with distance $r$ from the center of a uniformly charged spherical shell. [CBSE]

Solution:

1. Inside ($r < R$): $E = 0$.
2. Outside ($r \ge R$): $E = kQ / r^2$, so $E \propto 1/r^2$.
3. Plot: The graph is a zero line from $r=0$ to $r=R$, jumps to maximum at $R$, and then curves downward.

Question 6: Two point charges $+4\mu C$ and $+1\mu C$ are separated by a distance of 2 m in air. Find the point on the line joining charges where the net electric field is zero. [NEET]

Solution:

1. Setup: Let the point be $x$ from the $+4\mu C$ charge. Then distance from $+1\mu C$ is $(2-x)$.
2. Equate Fields: $k(4)/x^2 = k(1)/(2-x)^2$.
3. Simplify: $\sqrt{4}/x = \sqrt{1}/(2-x) \implies 2/x = 1/(2-x)$.
4. Solve: $4 - 2x = x \implies 3x = 4 \implies x = 1.33$ m.
5. Answer: 1.33 m from the $4\mu C$ charge.

Question 7: Why do electrostatic field lines not form closed loops? [CBSE]

Solution:

1. Nature of Field: Electrostatic fields are conservative in nature.
2. Work Done: If a field line formed a closed loop, the work done in moving a charge around that loop would be non-zero, which contradicts the property of conservative forces.
3. Reason: Lines always start at a positive charge (source) and end at a negative charge (sink).

Question 8: A spherical conductor of radius 10 cm has a charge of $3.2 \times 10^{-7}$ C distributed uniformly on its surface. What is the electric field at a point 15 cm from the center? [NEET]

Solution:

1. Context: $r = 0.15$ m, which is outside the sphere ($R = 0.1$ m).
2. Formula: $E = kQ/r^2$.
3. Calculate: $E = (9 \times 10^9) \times (3.2 \times 10^{-7}) / (0.15)^2$.
4. Result: $E = (28.8 \times 10^2) / 0.0225 = 1.28 \times 10^5$ N/C.

Question 9: What is the orientation of an electric dipole in a uniform electric field for (i) stable equilibrium and (ii) unstable equilibrium? [CBSE]

Solution:

1. Stable: The dipole moment $\vec{p}$ must be parallel to the electric field $\vec{E}$ (angle $\theta = 0^\circ$).
2. Unstable: The dipole moment $\vec{p}$ must be anti-parallel to the electric field $\vec{E}$ (angle $\theta = 180^\circ$).

Question 10: If the number of electric field lines leaving a closed surface is $8 \times 10^3$ and those entering it is $2 \times 10^3$, find the net charge enclosed by the surface. [CBSE]

Solution:

1. Net Flux: $\Phi_{net} = \text{Outward Flux} - \text{Inward Flux} = 8 \times 10^3 - 2 \times 10^3 = 6 \times 10^3$ N m$^2$/C.
2. Gauss's Law: $Q = \Phi \times \epsilon_0 = (6 \times 10^3) \times (8.854 \times 10^{-12})$.
3. Result: $Q \approx 5.31 \times 10^{-8}$ C.

Question 11: Calculate the work done in rotating a dipole of dipole moment $3 \times 10^{-8}$ C m from $0^\circ$ to $180^\circ$ in an external field of $10^3$ N/C. [NEET]

Solution:

1. Formula: $W = pE(\cos \theta_1 - \cos \theta_2)$.
2. Substitute: $W = (3 \times 10^{-8}) \times (10^3) \times (\cos 0^\circ - \cos 180^\circ)$.
3. Calculate: $W = (3 \times 10^{-5}) \times (1 - (-1)) = 6 \times 10^{-5}$ J.

Question 12: A metallic shell of radius $R$ has a charge $Q$. A point charge $q$ is placed at the center. Find the surface charge density on the inner and outer surfaces of the shell. [CBSE]

Solution:

1. Inner Surface: Due to induction, a charge of $-q$ appears on the inner surface. $\sigma_{inner} = -q / (4\pi R_{inner}^2)$.
2. Outer Surface: The total charge $Q$ redistributes. The outer surface gets the original charge $Q$ plus the induced $+q$. $\sigma_{outer} = (Q+q) / (4\pi R_{outer}^2)$.

Question 13: In a region, the electric field is $\vec{E} = 3/5 E_0 \hat{i} + 4/5 E_0 \hat{j}$. Find the flux through a rectangular area $S$ in the y-z plane. [NEET]

Solution:

1. Area Vector: An area in the y-z plane has its normal along the x-axis. So, $\vec{A} = S \hat{i}$.
2. Flux: $\Phi = \vec{E} \cdot \vec{A} = (3/5 E_0 \hat{i} + 4/5 E_0 \hat{j}) \cdot (S \hat{i})$.
3. Result: $\Phi = 3/5 E_0 S$.

Question 14: Define 'Electric Flux'. Is it a scalar or vector quantity? [CBSE]

Solution:

1. Definition: Electric flux is defined as the total number of electric field lines passing through a given area. Mathematically, it is the dot product of the electric field and the area vector.
2. Nature: It is a scalar quantity.

Question 15: Two identical metallic spheres A and B, each carrying a charge $q$, repel each other with force $F$. A third identical uncharged sphere C is touched to A, then to B, and finally removed. What is the new force between A and B? [CBSE]

Solution:

1. Step 1 (C touches A): $q_A = q/2, q_C = q/2$.
2. Step 2 (C touches B): $q_B = (q/2 + q)/2 = 3q/4$.
3. Step 3 (New Force): $F' = k (q/2)(3q/4) / r^2 = 3/8 (kq^2/r^2)$.
4. Result: $F' = 3/8 F$.

Question 16: An infinite plane sheet of charge has a surface charge density of $10^{-7}$ C/m$^2$. Calculate the electric field at a distance of 5 cm from it. [NEET]

Solution:

1. Analysis: The field near an infinite sheet is independent of distance $r$.
2. Formula: $E = \sigma / (2\epsilon_0)$.
3. Calculate: $E = 10^{-7} / (2 \times 8.854 \times 10^{-12})$.
4. Result: $E \approx 5.64 \times 10^3$ N/C.

Question 17: A charge $Q$ is divided into two parts $q$ and $Q-q$. What is the ratio $Q/q$ so that the force between them is maximum? [NEET]

Solution:

1. Logic: Force is maximum when the product of charges $q(Q-q)$ is maximum.
2. Differentiate: $d/dq (qQ - q^2) = Q - 2q = 0 \implies q = Q/2$.
3. Ratio: $Q/q = 2$.

Question 18: Why is it safe to be inside a car during lightning? [CBSE]

Solution:

1. Electrostatic Shielding: A car's body is metallic.
2. Inside Conductors: The electric field inside a hollow conductor is zero. The metal body channels the lightning's current safely to the ground.

Question 19: An electron and a proton are placed in the same uniform electric field. Compare the magnitude of the forces acting on them and their accelerations. [NEET]

Solution:

1. Force: $F = qE$. Since $|q_e| = |q_p|$, the magnitudes of forces are equal.
2. Acceleration: $a = F/m$. Since $m_p \approx 1836 m_e$, the acceleration of the electron is much higher than that of the proton.

Question 20: How many electrons should be removed from a body to give it a charge of $1.6 \mu C$? [CBSE]

Solution:

1. Formula: $n = Q/e$.
2. Values: $Q = 1.6 \times 10^{-6}$ C, $e = 1.6 \times 10^{-19}$ C.
3. Calculate: $n = (1.6 \times 10^{-6}) / (1.6 \times 10^{-19}) = 10^{13}$.
4. Result: $10^{13}$ electrons.

Question 21: Draw the electric field lines for a system of two equal positive charges. [CBSE]

Solution:

1. Concept: The lines emerge from both charges and repel each other. There is a neutral point at the center where $E=0$.

Question 22: If a body is earthed, what happens to its potential? [CBSE]

Solution:

1. Reference Point: The Earth is considered to be at zero electrical potential.
2. Effect: Earthing a body brings its potential to zero by allowing flow of electrons between the body and the Earth.

Question 23: The force between two point charges placed at a certain distance is $F$. If a brass plate is introduced between them, what will be the new force? [NEET]

Solution:

1. Dielectric Constant: Brass is a metal (conductor). For conductors, $K = \infty$.
2. Formula: $F_{medium} = F / K = F / \infty = 0$.
3. Answer: The force becomes Zero.

Question 24: A charge $q$ is placed at the center of the line joining two equal charges $Q$. Show that the system of three charges will be in equilibrium if $q = -Q/4$. [CBSE]

Solution:

1. Net Force on $Q$: For the system to be in equilibrium, the net force on the outer charge $Q$ must be zero.
2. Equation: $kQ^2/(2r)^2 + kQq/r^2 = 0$.
3. Solve: $Q^2/4r^2 + Qq/r^2 = 0 \implies Q/4 + q = 0 \implies q = -Q/4$.

Question 25: What is the work done in moving a test charge $q_0$ between two points on an equipotential surface? [CBSE]

Solution:

1. Concept: On an equipotential surface, $V_A = V_B$.
2. Formula: $W = q_0 (V_B - V_A)$.
3. Result: $W = q_0(0) = 0$. The work done is Zero.

Question 26: Two charges of magnitudes $+2\mu C$ and $-2\mu C$ are separated by a distance 2 cm. Find the dipole moment. [NEET]

Solution:

1. Values: $q = 2 \times 10^{-6}$ C, $2a = 0.02$ m.
2. Calculate: $p = q \times 2a = (2 \times 10^{-6}) \times 0.02 = 4 \times 10^{-8}$ C m.
3. Direction: From $-2\mu C$ to $+2\mu C$.

Question 27: A point charge $+q$ is placed at distance $r$ from an infinite line of charge with density $\lambda$. Find the force on the point charge. [NEET]

Solution:

1. Field of Line Charge: $E = \lambda / (2\pi\epsilon_0 r)$.
2. Force: $F = qE = q\lambda / (2\pi\epsilon_0 r)$.
3. Direction: Radially away from the line if $\lambda$ and $q$ are same sign.

Question 28: Can a body have a charge of $2.4 \times 10^{-19}$ C? [CBSE]

Solution:

1. Check Quantization: $n = Q/e = (2.4 \times 10^{-19}) / (1.6 \times 10^{-19}) = 1.5$.
2. Analysis: Since $n$ is not an integer, this charge is not possible.

Question 29: What is the angle between the electric dipole moment and the electric field at a point on the equatorial line? [NEET]

Solution:

1. Direction of p: Points from $-q$ to $+q$.
2. Direction of E: At the equatorial point, the net field is opposite to $\vec{p}$.
3. Angle: The angle is $180^\circ$.

Question 30: A charge $q$ is placed at one corner of a cube of side $a$. Find the flux through one face of the cube not touching the charge. [NEET]

Solution:

1. Total Flux through cube: The charge at the corner is shared by 8 cubes. Total flux through one cube is $q / 8\epsilon_0$.
2. Face Analysis: 3 faces meet at the corner where the charge is. Flux through them is zero (field lines are parallel to these surfaces).
3. Remaining Faces: The remaining 3 faces share the flux equally.
4. Result: $\Phi = (q / 8\epsilon_0) / 3 = q / 24\epsilon_0$.