Solutions for RBSE Class 12 Physics Chapter 2: Electrostatic Potential and Capacitance

Are you an RBSE Class 12 student looking to master Electrostatic Potential and Capacitance? This detailed guide provides complete, step-by-step solutions for Chapter 2 of your Physics textbook (based on the NCERT curriculum adopted by the Rajasthan Board of Secondary Education), designed for high performance in your board exams and competitive tests.

Chapter 2 Overview: Electrostatic Potential and Capacitance

Chapter 2 is a cornerstone of electrostatics, building upon the concepts of electric charges and fields from Chapter 1. Understanding these topics is crucial for solving both theoretical and numerical problems.

Key Topics Covered in this Chapter:

1. Electrostatic Potential: Definition, potential difference, and its relationship with the electric field (E=−drdV​).
2. Potential due to a Point Charge and a System of Charges: Derivation and numerical applications.
3. Electric Dipole Potential: Potential at axial and equatorial points.
4. Equipotential Surfaces: Properties and representation for various charge configurations.
5. Electrostatic Potential Energy: Energy of a two-charge system and an electric dipole in a uniform electric field.
6. Conductors in an Electrostatic Field: Key properties like zero electric field inside and the entire volume being equipotential.
7. Capacitors and Capacitance: Definition, principle, and capacitance of a parallel plate capacitor.
8. Effect of Dielectrics: Polarisation, dielectric strength, and the effect of placing a dielectric slab between capacitor plates.
9. Capacitor Combinations: Series and parallel grouping and calculating equivalent capacitance.
10. Energy Stored in a Capacitor: Formula and energy density.

Detailed NCERT Exercise Solutions for RBSE Class 12 Physics Chapter 2

Below are the solutions to the most important and frequently asked questions from the NCERT textbook exercises for your RBSE board preparation.

Section A: Electric Potential Problems

Q. 2.1: Electric Potential at Zero Point Two charges q1​=5×10−8 C and q2​=−3×10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero?

Solution Strategy: The net electric potential (V) at any point P due to a system of charges is the algebraic sum of the potentials due to individual charges. V=V1​+V2​=0. We must consider two cases: a point between the charges and a point outside the charges.

• Case 1: Point P between the charges. (Calculation shows a point at 10 cm from q1​).
• Case 2: Point P outside the charges. (Calculation shows a point at 40 cm from q1​ on the side of the smaller magnitude charge q2​).

Q. 2.2: Potential at the Centre of a Hexagon A regular hexagon of side 10 cm has a charge 5μC at each of its vertices. Calculate the potential at the centre of the hexagon.

Solution: For a regular hexagon, the distance of the centre (r) from each vertex is equal to the side length, r=10 cm=0.1 m. Since there are N=6 equal charges (q=5×10−6 C), the total potential (V) at the centre is:

V=N×(4πϵ0​1​rq​)V=6×(9×109 Nm2/C2)×0.1 m5×10−6 C​V=2.7×106 V

Section B: Capacitance and Energy Problems

Q. 2.5: Effect of Dielectric and Plate Separation on Capacitance A parallel plate capacitor with air between the plates has a capacitance of 8 pF. What will be the capacitance if the distance between the plates is reduced by half and the space between them is filled with a substance of dielectric constant K=6?

Solution Strategy:

1. Initial Capacitance (Air): C0​=dϵ0​A​=8 pF.
2. New Setup: New distance d′=d/2 and new dielectric K=6.
3. New Capacitance: C′=d′Kϵ0​A​=(d/2)Kϵ0​A​=2K(dϵ0​A​)=2KC0​. C′=2×6×8 pF=96 pF

Q. 2.6: Capacitor Combination (Series) Three capacitors, each of capacitance 9 pF, are connected in series. (a) What is the total capacitance of the combination? (b) What is the potential difference across each capacitor if the combination is connected to a 120 V supply?

Solution: (a) For capacitors in series: Ceq​1​=C1​1​+C2​1​+C3​1​=91​+91​+91​=93​=31​ pF−1.

Ceq​=3 pF

(b) In a series combination, the charge (Q) on each capacitor is the same.

Q=Ceq​×V=(3×10−12 F)×120 V=360×10−12 C=360 pC

The potential difference across each identical capacitor is V′=CQ​=9 pF360 pC​=40 V. (Verification: V1​+V2​+V3​=40 V+40 V+40 V=120 V).

RBSE Board Exam Strategy: Important Derivations & Questions

For high scores in the RBSE Class 12 Physics paper, focus on these critical areas:

Topic Type	Important Questions/Derivations
Derivation	Expression for electric potential due to an electric dipole.
Derivation	Expression for the capacitance of a parallel plate capacitor (with and without dielectric).
Derivation	Energy stored in a capacitor, U=21​CV2=2CQ2​.
Conceptual	Properties of Equipotential Surfaces and Electric Field lines.
Numerical	Problems on series and parallel combinations of capacitors.
Numerical	Calculating electric potential and potential energy for charge configurations (e.g., triangle, square).

How to Achieve a High Rank in Physics Exams

1. Conceptual Clarity: Ensure you fully understand the definitions and principles of electric potential, equipotential surfaces, and capacitance.
2. Formula Mastery: Create a dedicated list of all formulas from this chapter (Potential, Capacitance, Energy, Combinations) and memorize them.
3. Practice Numericals: The Rajasthan Board frequently asks numerical problems from this chapter. Practice all textbook examples and exercise questions multiple times.
4. Derivation Practice: Write out the derivations repeatedly. Pay attention to the diagrams and clear labelling of terms.

Chapter 2

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Mastering the solutions and concepts of RBSE Class 12 Physics Chapter 2 will significantly boost your confidence and marks in your board examinations.