Class 12 : Physics (English) – Chapter 2: Electrostatic Potential and Capacitance

EXPLANATION & SUMMARY

Introduction
Electrostatics is the study of electric charges at rest. In the previous chapter, we learned about electric charges and electric fields. In this chapter, we extend our understanding to electrostatic potential, the concept of potential energy, and the behavior of conductors in an electric field. We also explore capacitance, a measure of how much charge a system can store, and how capacitors work.

Electrostatic Potential
The electrostatic potential at a point in space is defined as the work done per unit positive test charge to bring it from infinity to that point, without any acceleration.
Mathematically:
V = W/q₀
Where:
V is the electrostatic potential
W is the work done in bringing the charge
q₀ is the test charge
It is a scalar quantity and its SI unit is volt (V).

Potential Due to a Point Charge
Let a charge q be located at a point in space. The potential at a distance r from the charge is:
V = (1/4πε₀) × (q/r)
This shows that the potential decreases with distance and is directly proportional to the magnitude of the charge.

Potential Due to an Electric Dipole
An electric dipole consists of two equal and opposite charges separated by a small distance. The potential at a point P at a distance r from the center of the dipole, making an angle θ with the axis, is:
V = (1/4πε₀) × (p cosθ / r²)
Where p = q × 2a is the dipole moment.

Potential Due to a System of Charges
If multiple charges q₁, q₂, q₃,… are present, the potential at a point due to all of them is:
V = V₁ + V₂ + V₃ + … = Σ (1/4πε₀) × (qᵢ / rᵢ)
Since potential is a scalar quantity, we add algebraically.

Equipotential Surfaces
Equipotential surfaces are those where the potential is constant everywhere. No work is done in moving a test charge on an equipotential surface.
Key Properties:
Electric field lines are perpendicular to equipotential surfaces.
For a point charge, equipotential surfaces are spherical in shape.
The closer the surfaces, the stronger the electric field.

Relationship Between Electric Field and Potential
The electric field is related to the rate of change of potential with distance:
E = -dV/dr
In vector form:
𝐄 = -∇V
This negative sign indicates that electric field points in the direction of decreasing potential.

Electrostatic Potential Energy
It is the work done in assembling a system of charges. For two point charges q₁ and q₂ separated by a distance r:
U = (1/4πε₀) × (q₁q₂ / r)
For a system of three or more charges, we calculate the energy for each pair and sum them.

Potential Energy in an External Field
A single charge q in an external field E has potential energy:
U = qV
A dipole in a uniform electric field has potential energy:
U = -p · E = -pE cosθ
Where θ is the angle between the dipole moment vector p and the electric field E.

Conductors in Electrostatic Equilibrium
Important Properties:
Electric field inside a conductor is zero.
Potential is constant throughout the conductor and on its surface.
Electric field just outside a conductor is perpendicular to its surface.
Charge resides only on the outer surface of a conductor.

Capacitance
Capacitance is the ability of a conductor to store charge.
Defined as:
C = q/V
Where:
C is the capacitance
q is the charge stored
V is the potential difference
SI Unit: Farad (F)
1 F = 1 C/V

Capacitance of an Isolated Spherical Conductor
For a spherical conductor of radius R:
C = 4πε₀R
It shows that capacitance is directly proportional to the radius of the sphere.

Parallel Plate Capacitor
Two large conducting plates separated by a small distance d, area A, facing each other, form a capacitor.
Capacitance:
C = ε₀A/d
This is the most basic capacitor design.

Capacitor with Dielectric
A dielectric is a non-conducting material inserted between the plates of a capacitor. It increases the capacitance by reducing the effective electric field.
New capacitance:
C = Kε₀A/d
Where:
K is the dielectric constant (>1)
The dielectric increases capacitance K times.

Series and Parallel Combination of Capacitors
Series Combination:
1/C = 1/C₁ + 1/C₂ + 1/C₃ + …
Total voltage divides, same charge on each.
Parallel Combination:
C = C₁ + C₂ + C₃ + …
Voltage same across all, charge divides.

Energy Stored in a Capacitor
When a capacitor is charged, it stores energy in the form of electrostatic potential energy.
Stored energy:
U = ½ CV²
Other forms:
U = ½ qV = q² / (2C)
This energy is stored in the electric field between the plates.

Energy Density of Electric Field
Energy stored per unit volume between the plates:
u = ½ ε₀E²
Where E is the electric field.

Van de Graaff Generator
This is a high-voltage electrostatic generator based on the principle of charging by conduction.
Working principle:
A moving belt transfers charge to a hollow metallic sphere.
Charges accumulate on the outer surface, achieving high potential.
Used in particle accelerators and nuclear experiments.

✍ SUMMARY (Approx. 300 Words)
Electrostatic potential (V) is the work done per unit positive test charge to bring it from infinity to a point in an electric field.
V = (1/4πε₀) × (q/r) for a point charge.
Electric dipole produces potential:
V = (1/4πε₀) × (p cosθ / r²)
Equipotential surfaces are surfaces with constant potential and are perpendicular to electric field lines.

The electric field is the negative gradient of the potential:
E = -dV/dr
Electrostatic potential energy is the work required to assemble a system of charges. For two charges:
U = (1/4πε₀)(q₁q₂/r)
A conductor in electrostatic equilibrium has zero electric field inside and constant potential throughout.

Capacitance (C) is the ability to store charge per unit voltage:
C = q/V
Capacitance of:
Isolated sphere: C = 4πε₀R
Parallel plates: C = ε₀A/d
With dielectric: C = Kε₀A/d

In series, capacitors share same charge; voltage adds:
1/C = 1/C₁ + 1/C₂ + …
In parallel, capacitors share same voltage; charge adds:
C = C₁ + C₂ + …
Energy stored: U = ½ CV²
Energy density: u = ½ ε₀E²
Van de Graaff generator is a device to generate very high voltages using a moving belt and a metallic dome.

This chapter bridges electric field concepts with potential energy and introduces storage elements like capacitors, essential in circuits and electrostatic devices.

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QUESTIONS FROM TEXTBOOK

Q 2.1: Zero Potential Points
Question: Two charges +5×10⁻⁸ C and –3×10⁻⁸ C are 0.16 m apart. Locate the point(s) on the line joining them where the electric potential is zero.
Answer and Discussion:
Let the +5×10⁻⁸ C charge be at x = 0 and the –3×10⁻⁸ C charge at x = 0.16 m. Let the point where potential is zero be at a distance r from the +5×10⁻⁸ C charge.
Total potential at that point:
V = k [5×10⁻⁸ / r – 3×10⁻⁸ / (0.16 – r)] = 0
Solving:
(5×10⁻⁸)/r = (3×10⁻⁸)/(0.16 – r)
5(0.16 – r) = 3r
0.8 – 5r = 3r
0.8 = 8r → r = 0.10 m or 10 cm
Another case: Point beyond the –3×10⁻⁸ C charge (r > 0.16 m)
5/(r) = 3/(r – 0.16)
5(r – 0.16) = 3r → 5r – 0.8 = 3r → 2r = 0.8 → r = 0.4 m or 40 cm
Thus, the points where potential is zero are at 10 cm and 40 cm from the +5×10⁻⁸ C charge.
Extended Concept: At these points, the algebraic sum of potentials due to both charges cancels out. Unlike electric field (a vector), potential is scalar and adds algebraically. Therefore, two points exist for opposite charges where the potential is zero—one between them and one outside.

Q 2.2: Potential at Center of Hexagon
Question: A regular hexagon has charges of +5 μC at each vertex. Side of hexagon is 10 cm. Find the potential at the center.
Answer and Discussion:
Distance from center to any vertex = side = 0.10 m
Total potential at center:
V = 6 × kq / r = 6 × (9×10⁹) × (5×10⁻⁶) / 0.10
= 6 × 45×10³ = 270×10³ V = 2.7×10⁵ V
Extended Concept: All six vertices contribute equally due to symmetry. Electric potential, being a scalar, adds up simply. This demonstrates the superposition principle for potential and shows how symmetry simplifies calculations.

Q 2.3: Equipotential Surface for a Dipole
Question: Two charges +2 μC and –2 μC are 6 cm apart.
(a) Identify an equipotential surface.
(b) What is the direction of the electric field at that surface?
Answer and Discussion:
(a) The perpendicular bisector of the dipole is an equipotential surface since every point on it is equidistant from both charges. The net potential at any point on this bisector is zero.
(b) The electric field is always perpendicular to equipotential surfaces and points from the positive charge to the negative charge.
Extended Concept: Equipotential surfaces are always perpendicular to electric field lines. In the case of a dipole, the mid-plane is at zero potential and forms an important basis for understanding molecular dipoles and electrostatic shielding.

Q 2.4: Electric Field Around Spherical Conductor
Question: A spherical conductor of radius 12 cm carries charge 1.6×10⁻⁷ C. Find electric field:
(a) Inside the sphere
(b) Just outside the surface
(c) At 18 cm from the center
Answer and Discussion:
(a) Inside a conductor: E = 0
(b) Outside surface (r = 0.12 m):
E = kQ / r² = (9×10⁹ × 1.6×10⁻⁷) / (0.12)²
= 1.0×10⁵ N/C
(c) At 18 cm = 0.18 m:
E = (9×10⁹ × 1.6×10⁻⁷) / (0.18)² = 4.4×10⁴ N/C
Extended Concept: This follows from Gauss’s Law. Inside a conductor, charges rearrange such that the net electric field is zero. Outside, the field behaves as if the entire charge is concentrated at the center.

Q 2.5: Capacitance with Dielectric and Plate Distance Change
Question: A parallel-plate capacitor has air and capacitance 8 pF. Distance is halved, and dielectric of k = 6 inserted. Find new capacitance.
Answer and Discussion:
Original: C ∝ 1/d
New: C’ = k × (C × 2) = 2kC = 2×6×8 = 96 pF
Extended Concept: Capacitance increases due to both decreased plate separation and the dielectric. Dielectrics reduce effective electric field, allowing more charge storage for the same voltage.

Q 2.6: Capacitors in Series
Question: Three capacitors of 9 pF each in series.
(a) Find equivalent capacitance
(b) Find voltage across each when connected to 120 V
Answer and Discussion:
(a) 1/Ceq = 1/9 + 1/9 + 1/9 = 3/9 → Ceq = 3 pF
(b) Q = Ceq × V = 3×10⁻¹² × 120 = 3.6×10⁻¹⁰ C
V across each: V = Q/C = 3.6×10⁻¹⁰ / 9×10⁻¹² = 40 V
Extended Concept: In series, charge is constant across capacitors. Voltage divides in inverse ratio to capacitance. This principle is used in high-voltage circuits where multiple capacitors share potential.

Q 2.7: Capacitors in Parallel
Question: Capacitors 2 pF, 3 pF, 4 pF in parallel.
(a) Find equivalent capacitance
(b) Find individual charges with 100 V supply
Answer and Discussion:
(a) Ceq = 2 + 3 + 4 = 9 pF
(b)
Q₁ = 2×10⁻¹² × 100 = 2.0×10⁻¹⁰ C
Q₂ = 3×10⁻¹² × 100 = 3.0×10⁻¹⁰ C
Q₃ = 4×10⁻¹² × 100 = 4.0×10⁻¹⁰ C
Extended Concept: Voltage remains constant across parallel branches, and total charge is the sum of individual charges. This configuration is ideal for increasing overall capacitance.

Q 2.8: Design of Parallel Plate Capacitor
Question: Plates of area 6×10⁻³ m², separated by 3 mm.
(a) Find capacitance
(b) Charge with 100 V supply
Answer and Discussion:
(a) C = ε₀A/d = (8.854×10⁻¹² × 6×10⁻³) / (3×10⁻³)
= 17.7 pF
(b) Q = CV = 17.7×10⁻¹² × 100 = 1.77×10⁻⁹ C
Extended Concept: Capacitance increases with plate area and decreases with separation. This is the working principle behind most commercial capacitors.

Q 2.9: Dielectric Insertion in Capacitor
Question: A 3 mm mica slab (k = 6) is inserted into the capacitor of Q2.8.
(a) With supply connected
(b) With supply disconnected
Answer and Discussion:
C’ = kC = 6 × 17.7 = 106 pF
(a) Supply connected (V = constant):
Q’ = C’V = 106×10⁻¹² × 100 = 1.06×10⁻⁸ C
E remains the same
(b) Supply disconnected (Q = constant):
V’ = Q / C’ = 1.77×10⁻⁹ / 106×10⁻¹² = 16.7 V
Extended Concept: With connected supply, charge increases. With disconnected supply, voltage decreases. The work done appears as mechanical energy or internal heating.

Q 2.10: Energy Stored in a Capacitor
Question: Capacitor of 12 pF connected to 50 V. Find energy stored.
Answer and Discussion:
U = ½ CV² = ½ × 12×10⁻¹² × (50)² = 1.5×10⁻⁸ J
Extended Concept: This energy is the work done in charging. It’s stored in the electric field and used in devices requiring sudden energy discharge like flashes and defibrillators.

Q 2.11: Energy Loss in Capacitor Sharing
Question: A 600 pF capacitor charged to 200 V is connected to an uncharged identical capacitor. Find energy lost.
Answer and Discussion:
Initial energy:
U₁ = ½ × 600×10⁻¹² × (200)² = 1.2×10⁻⁵ J
Total charge: Q = 600×10⁻¹² × 200 = 1.2×10⁻⁷ C
Final voltage: V = Q / (600 + 600)pF = 100 V
Final energy:
U₂ = ½ × 1200×10⁻¹² × (100)² = 6.0×10⁻⁶ J
Energy lost = U₁ – U₂ = 6.0×10⁻⁶ J
Extended Concept: Half the energy is always lost as heat due to redistribution current in the connecting wires. This is important in practical capacitor applications where loss minimization is needed.

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OTHER IMPORTANT QUESTIONS FOR EXAMS

(CBSE MODEL QUESTIONS PAPER)

ESPECIALLY MADE FROM THIS LESSON ONLY

Q1. Which of the following statements is true about electrostatic potential?
(A) It is a scalar quantity
(B) It is a vector quantity
(C) It depends only on charge and not on distance
(D) It is always positive
Answer: (A)

Q2. The SI unit of electric potential is:
(A) Joule
(B) Coulomb
(C) Volt
(D) Newton
Answer: (C)

Q3. The work done in moving a charge between two points having the same potential is:
(A) Maximum
(B) Minimum
(C) Zero
(D) Infinite
Answer: (C)

Q4. The potential due to a point charge at a distance r is:
(A) kq/r²
(B) kq/r
(C) kr/q
(D) kq²/r
Answer: (B)

Q5. Assertion (A): Electrostatic potential inside a charged conductor is constant.
Reason (R): The electric field inside a charged conductor is zero.
(A) Both A and R are true and R is the correct explanation of A
(B) Both A and R are true but R is not the correct explanation of A
(C) A is true but R is false
(D) A is false but R is true
Answer: (A)

Q6. Two equal charges are placed at a certain distance apart. The potential at the midpoint is:
(A) Zero
(B) Maximum
(C) Minimum
(D) Cannot be determined
Answer: (B)

Q7. Electric potential due to a dipole at any point on the equatorial line is:
(A) Zero
(B) Maximum
(C) Negative
(D) Positive
Answer: (A)

Q8. Which of the following has the dimensions of capacitance?
(A) [M⁻¹L⁻²T⁴A²]
(B) [ML²T⁻³A]
(C) [ML²T⁻²A⁻¹]
(D) [M⁻¹L⁻²T³A²]
Answer: (A)

Q9. Capacitance of a parallel plate capacitor is directly proportional to:
(A) Distance between plates
(B) Thickness of plates
(C) Area of plates
(D) Permittivity of vacuum only
Answer: (C)

Q10. When a dielectric slab is inserted between the plates of a charged capacitor, the potential difference:
(A) Increases
(B) Decreases
(C) Remains unchanged
(D) Becomes infinite
Answer: (B)

Q11. Which of the following is a correct unit of capacitance?
(A) Coulomb per meter
(B) Newton per meter
(C) Volt per coulomb
(D) Farad
Answer: (D)

Q12. A 2 µF capacitor is charged to 100 V. The energy stored in it is:
(A) 0.01 J
(B) 0.02 J
(C) 0.1 J
(D) 1.0 J
Answer: (A)
Explanation: U = ½ CV² = ½ × 2×10⁻⁶ × (100)² = 0.01 J

Q13. Which of the following increases the capacitance of a parallel plate capacitor?
(A) Increasing the distance between plates
(B) Decreasing the area of plates
(C) Inserting a dielectric slab
(D) Removing the dielectric
Answer: (C)

Q14. The electric potential at a point due to a dipole varies with distance r as:
(A) 1/r²
(B) 1/r³
(C) r
(D) r²
Answer: (B)

Q15. Very Short Answer:
What is the work done in moving a charge between two points on an equipotential surface?
Answer: Zero

Q16. Very Short Answer:
What is the relation between electric field and potential gradient?
Answer: E = –dV/dr

Q17. Case-Based MCQ:
A parallel plate capacitor is connected to a battery. A dielectric slab is now inserted between the plates.
Which of the following is correct?
(A) Capacitance decreases
(B) Charge remains constant
(C) Electric field increases
(D) Potential difference increases
Answer: (B)
Explanation: In battery-connected case, V is constant, so Q increases to keep C = Q/V.

Q18. Case-Based MCQ:
A spherical conductor of radius R is charged. The electric potential on its surface is V.
What will be the potential at its center?
(A) 0
(B) V
(C) 2V
(D) V/2
Answer: (B)


Section B: Q19–Q23 (2 marks each)
Q19. A 5 μC charge is placed at a point in space. What is the electric potential at a distance of 2 m from it in vacuum?
(Take k = 9 × 10⁹ Nm²/C²)
Answer:
V = kq/r
= (9 × 10⁹) × (5 × 10⁻⁶) / 2
= 45 × 10³ / 2 = 22.5 kV

Q20. Define equipotential surface. Sketch the equipotential surfaces due to a point charge.
Answer:
Equipotential surface is a surface on which the electric potential is the same at every point.
No work is done in moving a test charge over an equipotential surface.
Sketch: Concentric spheres around a point charge represent equipotential surfaces.

Q21. What is the energy stored in a capacitor of 10 μF when it is charged to 200 V?
Answer:
U = ½ CV²
= ½ × 10 × 10⁻⁶ × (200)²
= 5 × 10⁻⁶ × 40000 = 0.2 J

Q22. Two capacitors of capacitance 6 μF and 12 μF are connected in series across a 60 V battery. Find the potential difference across each.
Answer:
In series:
1/C_eq = 1/6 + 1/12 = (2+1)/12 = 1/4 ⇒ C_eq = 4 μF
Charge on each capacitor:
Q = C_eq × V = 4 × 10⁻⁶ × 60 = 240 μC
V₁ = Q/C₁ = 240/6 = 40 V
V₂ = Q/C₂ = 240/12 = 20 V
V₁ = 40 V, V₂ = 20 V

Q23. A capacitor is charged and then disconnected from the battery. A dielectric is then inserted between the plates. What happens to the capacitance, voltage, and energy?
Answer:
Capacitance increases (C → KC)
Voltage decreases (V → V/K)
Energy decreases (U → U/K)
(Because charge remains constant, but potential and energy decrease)

Section C: Q24–Q28 (3 marks each)
Q24. Derive the expression for the potential due to an electric dipole at a point on its axial line.
Answer:
Let dipole moment p = q × 2a, and point P be at distance r from center of dipole along the axis.
Potential at P:
V = (1/4πε₀) × [q/(r – a) – q/(r + a)]
= (1/4πε₀) × [2qa / (r² – a²)]
If r >> a, then r² – a² ≈ r² ⇒
V ≈ (1/4πε₀) × (2qa / r²) = (1/4πε₀) × (p cosθ / r²) for axial point (θ = 0)
So, V_axial = (1/4πε₀) × (p / r²)

Q25. A parallel plate capacitor with air between the plates has a capacitance of 10 pF. What will be the capacitance if the distance between the plates is reduced to half and dielectric of constant K = 6 is introduced?
Answer:
Original: C = ε₀A/d
New: C’ = K × ε₀A/(d/2) = 2K × C = 2×6×10 = 120 pF

Q26. Show that the energy stored in a charged capacitor is given by U = ½ CV².
Also write two other equivalent expressions for energy.
Answer:
From Q = CV,
Work done in charging from 0 to Q:
U = ∫₀^Q V dq = ∫₀^Q (q/C) dq = (1/2) Q²/C = ½ CV² = ½ QV
Other forms:
U = ½ CV² = ½ Q²/C = ½ QV

Q27. A capacitor of capacitance 5 μF is connected to a 100 V battery. It is then disconnected and a dielectric slab of dielectric constant 5 is inserted. Calculate the new potential difference and energy stored.
Answer:
Initial: Q = CV = 5 × 10⁻⁶ × 100 = 5 × 10⁻⁴ C
New capacitance: C’ = 5 × C = 25 μF
New voltage: V’ = Q/C’ = (5 × 10⁻⁴)/(25 × 10⁻⁶) = 20 V
Energy: U = ½ Q²/C’ = ½ × (5×10⁻⁴)² / (25×10⁻⁶) = 0.005 J
V’ = 20 V, U = 0.005 J

Q28. A 100 pF capacitor is charged to 250 V. Another uncharged 300 pF capacitor is then connected in parallel.
Find the final potential and energy loss in the system.
Answer:
Initial charge: Q = 100×10⁻¹² × 250 = 25×10⁻⁹ C
Total capacitance: C = 100 + 300 = 400 pF
Final voltage: V = Q/C = (25×10⁻⁹)/(400×10⁻¹²) = 62.5 V
Initial energy: U₁ = ½ × 100×10⁻¹² × (250)² = 3.125×10⁻⁶ J
Final energy: U₂ = ½ × 400×10⁻¹² × (62.5)² = 0.781×10⁻⁶ J
Loss = 3.125 – 0.781 = 2.344 μJ

Section D: Q29–Q31 (4 marks each – Case-based questions)
Q29. Case Study: A parallel plate capacitor is connected across a battery. After charging, the battery is disconnected and a dielectric of constant K is inserted between the plates.
Answer the following:
(a) What happens to the capacitance?
(b) What happens to the potential difference?
(c) What happens to the stored energy?
(d) Justify the change in energy.
Answer:
(a) Capacitance increases (C → KC)
(b) Voltage decreases (V → V/K)
(c) Energy decreases (U → U/K)
(d) Since battery is disconnected, charge remains constant. The insertion of dielectric increases capacitance, so potential drops, and energy (U = Q²/2C) also drops.

Q30. Case Study: An electric dipole consists of charges +q and –q separated by distance 2a. It is placed in a uniform electric field E.
Answer the following:
(a) What is the torque acting on the dipole?
(b) When is the torque maximum?
(c) What is the potential energy of dipole in the field?
(d) When is this potential energy minimum?
Answer:
(a) τ = pE sinθ
(b) Torque is maximum when θ = 90°, i.e., dipole is perpendicular to the field.
(c) U = –pE cosθ
(d) Minimum when θ = 0°, i.e., dipole aligned with field.

Q31. Case Study: A spherical conductor of radius R is charged to a potential V.
Answer the following:
(a) What is the electric field on its surface?
(b) What is the electric field inside the conductor?
(c) What is the potential at the center?
(d) How does potential vary with distance outside the sphere?
Answer:
(a) E = V/R
(b) E = 0 (electric field inside conductor is zero)
(c) V (same as surface, constant throughout)
(d) V(r) = kQ/r ⇒ varies inversely with distance r (like point charge)

Section E: Q32–Q35 (5 marks each – Long answer questions)

Q32. Derive an expression for the capacitance of a parallel plate capacitor. Explain how the capacitance changes with insertion of a dielectric slab.
Answer:
Let area of plates = A, separation = d.
Electric field between plates: E = σ/ε₀ = Q/(Aε₀)
Potential difference: V = Ed = Qd/(Aε₀)
Capacitance: C = Q/V = Aε₀/d
C = ε₀A/d
If dielectric (K) is inserted:
C’ = Kε₀A/d = KC
Hence, capacitance increases K times.

Q33. Three capacitors of 2 μF, 3 μF, and 6 μF are connected in (a) series, (b) parallel. Calculate the equivalent capacitance in both cases. Also find energy stored when connected across 12 V in both arrangements.
Answer:
(a) Series:
1/C = 1/2 + 1/3 + 1/6 = (3+2+1)/6 = 6/6 = 1 ⇒ C_s = 1 μF
U = ½ × 1×10⁻⁶ × (12)² = 72 μJ
(b) Parallel:
C_p = 2 + 3 + 6 = 11 μF
U = ½ × 11×10⁻⁶ × (12)² = 792 μJ

Q34. A capacitor of capacitance 20 μF is connected to a 220 V battery.
(a) Calculate the charge stored and energy stored.
(b) If this capacitor is disconnected and connected to another uncharged 20 μF capacitor, what will be the final voltage and energy lost?
Answer:
(a)
Q = CV = 20×10⁻⁶ × 220 = 4.4 × 10⁻³ C
U = ½ CV² = ½ × 20×10⁻⁶ × (220)² = 0.484 J
(b)
Total capacitance = 40 μF
Final V = Q / C = 4.4 × 10⁻³ / 40×10⁻⁶ = 110 V
Final U = ½ × 40×10⁻⁶ × (110)² = 0.242 J
Energy lost = 0.484 – 0.242 = 0.242 J

Q35. Derive an expression for the electrostatic potential energy stored in a system of two point charges q₁ and q₂ separated by a distance r in vacuum. Also discuss the nature of potential energy for like and unlike charges.
Answer:
Work done in bringing q₂ from infinity to distance r from q₁:
V₁ = kq₁/r
U = q₂V₁ = kq₁q₂/r
Thus,
U = (1/4πε₀) × (q₁q₂ / r)
If q₁ and q₂ are like charges, U is positive (repulsion).
If unlike charges, U is negative (attraction).

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