Chapter 2: Electrostatic Potential and Capacitance Case Study Questions | physics Class 12 CBSE

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Chapter 2: Electrostatic Potential and Capacitance Case Study Questions | physics Class 12 CBSE | ByteNotes.in

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Riya is a Class 12 student studying electrostatics. Her teacher demonstrates an experiment: a positively charged rod is brought near a neutral metallic sphere. The sphere is then earthed briefly while the charged rod is still nearby, and finally the earth connection is removed before the rod is taken away. The sphere is found to be negatively charged. Her teacher explains that this process is called electrostatic induction. She also notes that if the sphere were made of glass (an insulator) instead of metal, the effect would be different — some polarisation would occur but no charge would be transferred to earth. The teacher connects this to the concept of electrostatic potential, noting that when earthed, the potential of the sphere becomes zero.

Question 1: Why does the sphere acquire a negative charge after the above process, even though it never touched the charged rod?

When the positive rod is near the sphere, free electrons in the metal are attracted toward the rod's side and positive charges move to the far side. When earthed, electrons flow from earth to the sphere to reduce its potential to zero, giving it excess negative charge.
When the earth connection is removed (with rod still present), this excess negative charge is trapped. Removing the rod then allows this negative charge to redistribute over the sphere — the sphere remains negatively charged by induction.

Question 2: Why would the effect be different if the sphere were made of glass instead of metal?

Glass is a dielectric (insulator) with no free charge carriers. Induction requires free charges to move and be transferred to earth — this cannot happen in glass.
In glass, only polarisation (bound charge displacement) occurs: molecules develop induced dipole moments. No net charge can be transferred to earth, so the glass sphere cannot be charged by induction.

Question 3: When the metallic sphere is earthed with the rod nearby, explain in terms of electrostatic potential why electrons flow from earth to the sphere. Also explain why the potential of the sphere equals zero when earthed.

The positive rod raises the potential of the near side of the sphere and lowers it on the far side. The overall potential of the sphere (average) is positive because induced positive charges on the far side have higher potential due to proximity to the positive rod.
Earth is at zero potential (defined reference). When the sphere's potential > 0 (due to the rod's influence), electrons flow from the lower potential earth to the higher potential sphere to equalise potentials — the sphere's potential becomes zero.
This flow of electrons adds negative charge to the sphere, which stays even after the earth connection is removed. The process demonstrates that charge flows in response to potential difference, always from lower to higher potential for electrons.

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Case Set 2

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A physics teacher is demonstrating capacitors in the laboratory. She has a parallel plate capacitor with air between plates (plate area A = 0.02 m², separation d = 2 mm). She connects it to a 100 V battery, then disconnects it. She then slides a mica sheet (dielectric constant K = 6) of the same area and thickness 2 mm between the plates. The students observe the voltage reading on an electrometer connected across the capacitor plates drops significantly. The teacher asks the students to calculate the original capacitance, the new capacitance, and the energy stored before and after the dielectric insertion, and to reason about the energy change.

Question 1: Calculate the original capacitance C₀ of the air capacitor and the charge Q stored when connected to 100 V.

C₀ = ε₀A/d = (8.85 × 10⁻¹² × 0.02)/( 2 × 10⁻³) = 8.85 × 10⁻¹¹ F = 88.5 pF.
Charge Q = C₀V = 88.5 × 10⁻¹² × 100 = 8.85 × 10⁻⁹ C = 8.85 nC. This charge remains constant after disconnection from battery.

Question 2: What is the new voltage across the capacitor after the mica sheet is inserted? Explain why it changes.

New capacitance C = KC₀ = 6 × 88.5 = 531 pF. Since Q is constant (disconnected), new voltage V' = Q/C = 8.85 × 10⁻⁹/(531 × 10⁻¹²) = 100/6 ≈ 16.7 V.
The voltage drops because mica polarises in the electric field, creating an opposing internal field that reduces the net field between plates. Since V = Ed, the potential difference decreases by factor K = 6.

Question 3: Calculate the energy stored before and after dielectric insertion. Where does the energy difference go? Explain the physical mechanism.

Initial energy: U_i = (1/2)C₀V² = (1/2)(88.5 × 10⁻¹²)(100²) = 4.425 × 10⁻⁷ J. Final energy: U_f = Q²/(2C) = (8.85 × 10⁻⁹)²/(2 × 531 × 10⁻¹²) = 7.37 × 10⁻⁸ J = U_i/K.
Energy decreased: ΔU = U_i − U_f = 4.425 × 10⁻⁷ − 7.37 × 10⁻⁸ ≈ 3.69 × 10⁻⁷ J. The decrease equals U_i(1 − 1/K).
This energy is not lost — it is converted to mechanical work done by the electrostatic force pulling the dielectric slab into the capacitor. The system spontaneously lowers its potential energy by drawing in the dielectric, analogous to a spring releasing potential energy.

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An electronics engineer is designing a filter circuit and needs to combine capacitors. She has capacitors: C₁ = 2 μF, C₂ = 4 μF, and C₃ = 6 μF. She needs to achieve a specific equivalent capacitance of 3 μF for one circuit and 12 μF for another circuit using these three capacitors. She also needs to ensure that no individual capacitor exceeds its maximum rated voltage of 200 V. She connects the 3 μF configuration to a 600 V supply and the 12 μF configuration to a 100 V supply and calculates the charge and energy stored in each.

Question 1: Which combination of C₁, C₂, C₃ gives 3 μF and which gives 12 μF?

For 12 μF: connect all three in parallel. C_parallel = 2 + 4 + 6 = 12 μF. All share voltage 100 V (< 200 V limit — safe).
For 3 μF: C₃ in series with [C₁ || C₂]: C₁||C₂ = 6 μF; series with C₃ (6 μF): 1/C = 1/6 + 1/6 = 2/6, C = 3 μF ✓.

Question 2: For the 3 μF configuration connected to 600 V, find the charge on C₃ and verify no capacitor exceeds 200 V.

Config: C₃ (6 μF) in series with [C₁ || C₂] = 6 μF. Total C = 3 μF. Charge Q = CV = 3 × 10⁻⁶ × 600 = 1.8 × 10⁻³ C = 1.8 mC. In series, C₃ and the parallel group each carry Q = 1.8 mC.
Voltage across C₃: V₃ = Q/C₃ = 1.8 × 10⁻³/(6 × 10⁻⁶) = 300 V — exceeds 200 V limit! This combination would damage C₃. Voltage across C₁||C₂: 600 − 300 = 300 V — also exceeds the limit. The 600 V supply is unsuitable for these rated capacitors.

Question 3: For the 12 μF parallel combination at 100 V, find the charge on each capacitor and total energy stored. Compare with energy if same charge were stored in only C₁.

Parallel: each capacitor at 100 V. Q₁ = 200 μC; Q₂ = 400 μC; Q₃ = 600 μC. Total Q = 1200 μC. Energy: U = (1/2)(12 × 10⁻⁶)(100²) = 60 mJ.
If same Q = 1200 μC stored in C₁ = 2 μF alone: V = 600 V. Energy: U = Q²/(2C₁) = (1.2 × 10⁻³)²/(4 × 10⁻⁶) = 360 mJ.
Parallel combination stores same charge at 6 times less energy (60 vs 360 mJ) because voltage is 6 times lower. Energy ∝ V² for same Q — parallel combination is far more energy-efficient for a given charge storage.

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During a thunderstorm, lightning occurs when the electric field in the atmosphere between a negatively charged cloud base and the positively charged ground exceeds the dielectric strength of air (approximately 3 × 10⁶ V/m). The cloud-ground system acts roughly like a parallel plate capacitor. A scientist measures the potential difference between the cloud base (height ~500 m above ground) and the ground to be about 10⁸ V just before a lightning strike. She is investigating whether a lightning conductor (a sharp metallic rod on a building connected to ground by a thick conductor) prevents lightning by safely discharging the cloud or by diverting the strike.

Question 1: Using the parallel plate capacitor model, estimate the electric field between the cloud base and the ground. Does this explain when lightning occurs?

Electric field E = V/d = 10⁸ V / 500 m = 2 × 10⁵ V/m. This is below the dielectric strength of air (3 × 10⁶ V/m), so the uniform model alone does not explain lightning.
In reality, the field is highly non-uniform and is greatly enhanced near pointed objects (trees, buildings, rod tips) where surface charge density σ is very high (high curvature). Local field can exceed 3 × 10⁶ V/m even when the average field is lower, initiating lightning.

Question 2: Explain the electrostatic principle behind why a lightning conductor has a sharp pointed tip rather than a rounded end.

For a conductor, the surface charge density σ is highest where the radius of curvature is smallest. At a sharp point, the curvature radius approaches zero, making σ very large.
Since E = σ/ε₀ at the conductor surface, the electric field at the tip is extremely high. This strong field ionises the surrounding air, creating a conducting path (corona discharge) that allows the cloud's charge to leak away safely to earth — preventing a destructive lightning strike.

Question 3: Model the cloud-ground system as a capacitor of capacitance 0.05 F with potential difference 10⁸ V. Calculate energy stored and charge. After lightning discharge (assuming 80% of charge neutralises), find the remaining energy. Comment on the physics.

Energy stored: U = (1/2)CV² = (1/2)(0.05)(10⁸)² = 2.5 × 10¹⁴ J. Charge: Q = CV = 0.05 × 10⁸ = 5 × 10⁶ C. (These are hypothetical order-of-magnitude estimates.)
After 80% charge neutralises: remaining Q' = 0.2Q = 10⁶ C. New voltage V' = Q'/C = 10⁶/0.05 = 2 × 10⁷ V. Remaining energy: U' = Q'²/(2C) = (10⁶)²/(0.1) = 10¹³ J = U/25.
The energy drops to 4% of original (factor of 25 = 1/0.2²) while charge drops to 20%. Energy ∝ Q² — neutralising most charge rapidly dissipates almost all energy as heat, light, sound, and electromagnetic radiation of the lightning bolt.

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Case Set 5

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A student is studying the electrostatics of conductors using a hollow metallic sphere of radius 10 cm, mounted on an insulating stand. She places a charge of +5 μC on the sphere. She then uses a proof plane (a small flat conductor on an insulating handle) and an electroscope to map the charge distribution. She finds that all charge resides on the outer surface, the electric field inside is exactly zero, and the potential is the same everywhere inside as on the surface. She then introduces a small positive charge q₀ = +1 μC inside the hollow sphere without touching the walls, and finds that charges redistribute on the inner and outer surfaces of the sphere.

Question 1: What is the electric potential at the surface and at the centre of the hollow sphere (radius 10 cm) when it carries only +5 μC, before the internal charge is introduced?

Potential at surface: V_surface = kQ/R = (9 × 10⁹ × 5 × 10⁻⁶)/0.10 = 4.5 × 10⁵ V.
Potential at centre (and everywhere inside): V_inside = V_surface = 4.5 × 10⁵ V. Since E = 0 inside the conductor, potential is constant throughout the interior, equal to its surface value.

Question 2: When the charge q₀ = +1 μC is placed inside the hollow sphere, what charges appear on the inner and outer surfaces of the sphere? Explain.

The +1 μC charge inside induces −1 μC on the inner surface (by Gauss's law: E = 0 inside the conductor material requires zero net enclosed charge, so inner surface must have −1 μC to cancel q₀).
By charge conservation: sphere originally had +5 μC. Inner surface has −1 μC, so outer surface has +5 − (−1) = +6 μC. The external field is as if a total charge +6 μC is at the centre.

Question 3: After q₀ is placed inside, calculate the new potential at the outer surface (R = 10 cm) and at a point 20 cm from the centre. Is the potential inside the cavity still constant? Explain.

Outer surface potential: external field behaves as if +6 μC is at centre. V_surface = k(6 × 10⁻⁶)/0.10 = 9 × 10⁹ × 6 × 10⁻⁶/0.10 = 5.4 × 10⁵ V.
At r = 20 cm: V = k(6 × 10⁻⁶)/0.20 = 2.7 × 10⁵ V. (Outer field behaves as point charge +6 μC.)
Inside the conductor shell E is still zero, so the shell material is at constant potential 5.4 × 10⁵ V. However, inside the hollow cavity, the potential is NOT uniform — contributions from q₀ and inner surface charges vary spatially. The cavity is NOT shielded when charges are present inside it.

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Chapter 2: Electrostatic Potential and Capacitance | Case Study Questions

Passage

Question 1: Why does the sphere acquire a negative charge after the above process, even though it never touched the charged rod?

Question 2: Why would the effect be different if the sphere were made of glass instead of metal?

Question 3: When the metallic sphere is earthed with the rod nearby, explain in terms of electrostatic potential why electrons flow from earth to the sphere. Also explain why the potential of the sphere equals zero when earthed.

Passage

Question 1: Calculate the original capacitance C₀ of the air capacitor and the charge Q stored when connected to 100 V.

Question 2: What is the new voltage across the capacitor after the mica sheet is inserted? Explain why it changes.

Question 3: Calculate the energy stored before and after dielectric insertion. Where does the energy difference go? Explain the physical mechanism.

Passage

Question 1: Which combination of C₁, C₂, C₃ gives 3 μF and which gives 12 μF?

Question 2: For the 3 μF configuration connected to 600 V, find the charge on C₃ and verify no capacitor exceeds 200 V.

Question 3: For the 12 μF parallel combination at 100 V, find the charge on each capacitor and total energy stored. Compare with energy if same charge were stored in only C₁.

Passage

Question 1: Using the parallel plate capacitor model, estimate the electric field between the cloud base and the ground. Does this explain when lightning occurs?

Question 2: Explain the electrostatic principle behind why a lightning conductor has a sharp pointed tip rather than a rounded end.

Question 3: Model the cloud-ground system as a capacitor of capacitance 0.05 F with potential difference 10⁸ V. Calculate energy stored and charge. After lightning discharge (assuming 80% of charge neutralises), find the remaining energy. Comment on the physics.

Passage

Question 1: What is the electric potential at the surface and at the centre of the hollow sphere (radius 10 cm) when it carries only +5 μC, before the internal charge is introduced?

Question 2: When the charge q₀ = +1 μC is placed inside the hollow sphere, what charges appear on the inner and outer surfaces of the sphere? Explain.

Question 3: After q₀ is placed inside, calculate the new potential at the outer surface (R = 10 cm) and at a point 20 cm from the centre. Is the potential inside the cavity still constant? Explain.

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