PYQ Masterclass: Boards & NEET

High-Yield Exam Tips & Trends

For Electrostatic Potential and Capacitance, examiner trends show a heavy focus on the behavior of capacitors when dielectrics are introduced, and the relationship between Electric Field and Potential.

Core Focus Areas:

1. The Dielectric Table: Memorize the state of $Q, V, E, C,$ and $U$ when a battery is connected vs. disconnected. This is the single most repeated topic in NEET and CBSE Section C/D.
2. Equipotential Geometry: Be prepared to draw surfaces for a point charge, a dipole, and two like charges. Remember: field lines are always perpendicular to these surfaces.
3. Energy Loss: In competitive exams, the energy loss formula $$\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}$$ is a major time-saver.
4. $E = -dV/dr$: Numerical problems involving differentiation or integration of potential are common in JEE and IAT.

Question 1: Can two equipotential surfaces intersect each other? Give reason. [CBSE]

Solution:

1. Statement: No, two equipotential surfaces can never intersect.
2. Reasoning: Each surface represents a specific value of electric potential. If they intersected, the point of intersection would have two different values of potential, which is impossible.
3. Field implication: Since the electric field is perpendicular to an equipotential surface, an intersection would also imply two directions of electric field at the same point, which is physically impossible.

Question 2: What is the work done in moving a test charge $q$ through a distance of 5 cm along the equatorial line of an electric dipole? [CBSE]

Solution:

1. Concept: For an electric dipole, the potential at any point on the equatorial line is zero.
2. Potential difference: Since $V = 0$ everywhere on this line, $$\Delta V = 0 - 0 = 0$$
3. Work done: $$W = q \Delta V = q(0) = 0$$
4. Conclusion: The work done is zero.

Question 3: A parallel plate capacitor with air between the plates is charged to a potential $V$. If the plates are moved farther apart, how does the capacitance change? [CBSE]

Solution:

1. Formula: $$C = \frac{\epsilon_0 A}{d}$$
2. Relation: Capacitance is inversely proportional to plate separation: $$C \propto \frac{1}{d}$$
3. Conclusion: If the distance $d$ increases, the capacitance decreases.

Question 4: A 10 $\mu$F capacitor is charged to 50 V. How much energy is stored in it? [NEET]

Solution:

1. Given: $$C = 10 \times 10^{-6} \text{ F}, \quad V = 50 \text{ V}$$
2. Formula: $$U = \frac{1}{2}CV^2$$
3. Calculation: $$U = 0.5 \times (10 \times 10^{-6}) \times (50)^2$$ $$U = 5 \times 10^{-6} \times 2500$$ $$U = 12500 \times 10^{-6} = 0.0125 \text{ J}$$
4. Final Answer: $$U = 1.25 \times 10^{-2} \text{ J}$$

Question 5: Why must the electrostatic field be normal to the surface of a conductor? [CBSE]

Solution:

1. Assume otherwise: Suppose the field had a tangential component along the surface.
2. Consequence: This tangential component would exert a force on free electrons on the surface and cause them to move.
3. Contradiction: In electrostatic equilibrium, charges must remain at rest.
4. Conclusion: Therefore, no tangential component can exist, and the electric field must be perpendicular to the surface.

Question 6: Two capacitors of 2 $\mu$F and 3 $\mu$F are connected in series. Find the equivalent capacitance. [CBSE]

Solution:

1. Formula: $$\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2}$$
2. Substitute: $$\frac{1}{C_s} = \frac{1}{2} + \frac{1}{3} = \frac{3+2}{6} = \frac{5}{6}$$
3. Invert: $$C_s = \frac{6}{5} = 1.2 \mu \text{F}$$
4. Final Answer: $$C_s = 1.2 \mu \text{F}$$

Question 7: A dielectric slab ($K=4$) is inserted into an isolated charged capacitor. What happens to the potential difference? [NEET]

Solution:

1. Condition: Since the capacitor is isolated, charge $Q$ remains constant.
2. Capacitance after insertion: $$C' = KC = 4C$$
3. Using $$V = \frac{Q}{C}$$ the new potential becomes: $$V' = \frac{Q}{C'} = \frac{Q}{4C} = \frac{V}{4}$$
4. Conclusion: The potential difference decreases by a factor of 4.

Question 8: Define the term 'potential gradient'. [CBSE]

Solution:

1. Definition: Potential gradient is the rate of change of electric potential with respect to distance.
2. Mathematical form: $$\frac{dV}{dr}$$
3. Relation to electric field: $$E = -\frac{dV}{dr}$$
4. SI unit: Volt per meter (V/m).

Question 9: Three charges $+q, -2q, +q$ are kept at the vertices of an equilateral triangle. What is the net potential at the centroid? [JEE Main]

Solution:

1. Symmetry: The distance from each vertex to the centroid is the same, say $r$.
2. Total potential: $$V_{net} = \frac{k}{r}(q_1 + q_2 + q_3)$$
3. Sum of charges: $$q + (-2q) + q = 0$$
4. Final Answer: $$V_{net} = 0$$

Question 10: How does the energy density of a capacitor change if the electric field between its plates is tripled? [IAT]

Solution:

1. Formula: $$u = \frac{1}{2} \epsilon_0 E^2$$
2. Relation: $$u \propto E^2$$
3. If the field becomes $3E$, then: $$u' \propto (3E)^2 = 9E^2$$
4. Conclusion: The energy density becomes 9 times.

Question 11: Plot a graph showing the variation of electric potential $V$ with distance $r$ from the center of a charged conducting sphere. [CBSE]

Solution:

1. Inside the sphere ($r < R$): Potential remains constant: $$V = \frac{kQ}{R}$$
2. Outside the sphere ($r > R$): Potential decreases as: $$V = \frac{kQ}{r}$$
3. Graph description: A horizontal line from $r=0$ to $r=R$, followed by a decreasing hyperbolic curve for $r>R$.

Question 12: An electron is accelerated through a potential difference of 1000 V. What is its kinetic energy in eV? [NEET]

Solution:

1. Concept: The energy gained by a charge moving through a potential difference is: $$W = qV$$
2. Definition of eV: One electron volt is the energy gained by an electron when it moves through 1 volt.
3. Therefore, through 1000 V, the electron gains: $$1000 \text{ eV}$$
4. Final Answer: $$KE = 1000 \text{ eV}$$

Question 13: What is the capacitance of an isolated spherical conductor of radius $R$? [CBSE]

Solution:

1. Potential of the sphere: $$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$$
2. Capacitance formula: $$C = \frac{Q}{V}$$
3. Substitute: $$C = \frac{Q}{\frac{1}{4\pi\epsilon_0} \frac{Q}{R}}$$
4. Therefore, $$C = 4\pi\epsilon_0 R$$
5. Final Answer: $$C = 4\pi\epsilon_0 R$$

Question 14: A dipole moment $p$ is placed along the direction of a uniform field $E$. Find the work required to rotate it by $90^{\circ}$. [CBSE]

Solution:

1. Initial angle: $$\theta_1 = 0^{\circ}$$
2. Final angle: $$\theta_2 = 90^{\circ}$$
3. Work formula: $$W = pE(\cos\theta_1 - \cos\theta_2)$$
4. Substitute: $$W = pE(\cos 0^{\circ} - \cos 90^{\circ}) = pE(1 - 0) = pE$$
5. Final Answer: $$W = pE$$

Question 15: Why is the potential constant inside a hollow charged conductor? [CBSE]

Solution:

1. Key fact: The electric field inside a conductor in electrostatic equilibrium is zero.
2. Relation: $$E = -\frac{dV}{dr}$$
3. If $$E = 0$$, then: $$\frac{dV}{dr} = 0$$
4. A zero derivative means potential does not change with position.
5. Conclusion: The potential is constant throughout the interior.

Question 16: Two metallic spheres of different radii are given the same charge. Which will be at a higher potential? [NEET]

Solution:

1. Formula: $$V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R}$$
2. For fixed charge $Q$, $$V \propto \frac{1}{R}$$
3. Therefore, the smaller sphere has the higher potential.
4. Final Answer: The smaller sphere will be at a higher potential.

Question 17: If a mica sheet is inserted between the plates of a charged capacitor while the battery is still connected, how does the charge on the plates change? [CBSE]

Solution:

1. Battery connected: Potential difference remains constant.
2. With dielectric: Capacitance increases: $$C' = KC$$
3. Since $$Q = CV$$ and $V$ is constant, charge increases in the same ratio.
4. Conclusion: The charge increases by a factor of $K$.

Question 18: What is the shape of equipotential surfaces for an infinite line charge? [CBSE]

Solution:

1. An infinite line charge has cylindrical symmetry.
2. Therefore, all points at the same radial distance from the line are at the same potential.
3. Final Answer: The equipotential surfaces are concentric cylindrical surfaces with the line charge as the common axis.

Question 19: A 2 $\mu$F capacitor is charged to 10 V and another 4 $\mu$F capacitor is charged to 20 V. They are connected in parallel. Find the common potential. [NEET]

Solution:

1. Formula for common potential: $$V = \frac{C_1V_1 + C_2V_2}{C_1 + C_2}$$
2. Substitute: $$V = \frac{(2 \times 10) + (4 \times 20)}{2 + 4} = \frac{20 + 80}{6}$$
3. Therefore, $$V = \frac{100}{6} \approx 16.67 \text{ V}$$
4. Final Answer: $$V \approx 16.67 \text{ V}$$

Question 20: The energy stored in a capacitor is 4 J. If the voltage is doubled, what will be the new energy? [CBSE]

Solution:

1. Formula: $$U = \frac{1}{2}CV^2 \Rightarrow U \propto V^2$$
2. If voltage becomes $2V$, energy becomes: $$(2)^2 = 4$$ times the original.
3. Therefore, $$U' = 4 \times 4 = 16 \text{ J}$$
4. Final Answer: $$16 \text{ J}$$

Question 21: Draw equipotential surfaces for a uniform electric field. [CBSE]

Solution:

1. In a uniform electric field, potential changes only along the field direction.
2. Therefore, surfaces of constant potential are perpendicular to the field.
3. Final Answer: Equipotential surfaces are equally spaced parallel planes perpendicular to the electric field lines.

Question 22: Define dielectric strength. [CBSE]

Solution:

1. Definition: Dielectric strength is the maximum electric field a dielectric can withstand without electrical breakdown.
2. Units: V/m or kV/mm.
3. Significance: It determines the maximum safe electric field or voltage before the material becomes conducting.

Question 23: A point charge $+q$ is placed at the center of a cube. What is the potential at one of the corners? [JEE Main]

Solution:

1. In a cube of side $L$, the distance from center to a corner is: $$r = \frac{\sqrt{3}}{2}L$$
2. Potential at the corner: $$V = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$$
3. Substitute $r$: $$V = \frac{1}{4\pi\epsilon_0} \frac{q}{\sqrt{3}L/2} = \frac{q}{2\pi\epsilon_0 \sqrt{3}L}$$
4. Final Answer: $$V = \frac{q}{2\pi\epsilon_0 \sqrt{3}L}$$

Question 24: What happens to the capacitance of a parallel plate capacitor when the area of the plates is halved? [CBSE]

Solution:

1. Formula: $$C = \frac{\epsilon_0 A}{d}$$
2. Since capacitance is directly proportional to area, $$C \propto A$$
3. If area becomes $A/2$, capacitance also becomes $C/2$.
4. Final Answer: The capacitance is halved.

Question 25: Work done in moving a test charge $q$ between two points A and B is zero. Does this mean A and B must be on the same equipotential surface? [CBSE]

Solution:

1. Work done is: $$W = q(V_B - V_A)$$
2. If $$W = 0$$ and $$q \neq 0$$, then: $$V_A = V_B$$
3. Points having the same potential lie on the same equipotential surface.
4. Final Answer: Yes, A and B can be taken as lying on the same equipotential surface.

Question 26: A 500 $\mu$F capacitor is charged at a steady rate of 100 $\mu$C/s. How long will it take to raise the potential difference to 10 V? [NEET]

Solution:

1. Required charge: $$Q = CV = 500 \times 10^{-6} \times 10 = 5 \times 10^{-3} \text{ C}$$
2. Charging rate: $$I = 100 \times 10^{-6} \text{ C/s}$$
3. Time required: $$t = \frac{Q}{I} = \frac{5 \times 10^{-3}}{100 \times 10^{-6}} = 50 \text{ s}$$
4. Final Answer: $$t = 50 \text{ s}$$

Question 27: Mention two properties of polar dielectrics. [CBSE]

Solution:

1. Their molecules possess a permanent dipole moment even in the absence of an external electric field.
2. In an external electric field, they polarize mainly by alignment of these permanent dipoles.

Question 28: What is the force between two plates of a capacitor with charge $Q$ and area $A$? [JEE Main]

Solution:

1. Field due to one plate: $$E = \frac{\sigma}{2\epsilon_0} = \frac{Q}{2A\epsilon_0}$$
2. Force on the other plate: $$F = QE = Q \cdot \frac{Q}{2A\epsilon_0}$$
3. Therefore, $$F = \frac{Q^2}{2A\epsilon_0}$$
4. Final Answer: $$F = \frac{Q^2}{2A\epsilon_0}$$

Question 29: In which direction does the electric potential decrease most rapidly? [CBSE]

Solution:

1. The relation between field and potential is: $$E = -\frac{dV}{dr}$$
2. Therefore, potential decreases most rapidly in the direction of the electric field.
3. Final Answer: Along the direction of the electric field.

Question 30: What is the net charge on a charged capacitor? [CBSE]

Solution:

1. The two plates carry equal and opposite charges: $$+Q \text{ and } -Q$$
2. Net charge: $$+Q + (-Q) = 0$$
3. Final Answer: The net charge on the capacitor is zero.