A space probe carries two identical parallel-plate capacitors, C₁ and C₂, each with plate area A = 4×10⁻² m² and plate separation d = 2×10⁻³ m. On Earth, both are connected in series across a 120 V battery and fully charged; the battery is then disconnected. During the mission, the probe enters a region of a planet's atmosphere and the space between the plates of C₁ only is completely filled with a dielectric of relative permittivity (dielectric constant) K = 5. The probe's onboard computer monitors voltage changes across individual capacitors to analyse the dielectric properties of the atmosphere.
A space probe carries two identical parallel-plate capacitors, C₁ and C₂, each with plate area A = 4×10⁻² m² and plate separation d = 2×10⁻³ m. On Earth, both are connected in series across a 120 V battery and fully charged; the battery is then disconnected. During the mission, the probe enters a region of a planet's atmosphere and the space between the plates of C₁ only is completely filled with a dielectric of relative permittivity (dielectric constant) K = 5.
(i) Calculate the capacitance of each capacitor before the dielectric is inserted. (ε₀ = 8.854×10⁻¹² C² N⁻¹ m⁻²)
(ii) When the two capacitors were connected in series across 120 V (before dielectric insertion), find the charge Q stored on the combination and the energy U₀ stored in the series combination.
(iii) After the battery is disconnected and the dielectric (K = 5) is inserted into C₁, find the new equivalent capacitance of the series combination.
(iv) The charge Q on the isolated series combination remains conserved after the dielectric is inserted. Find the new voltage across C₂ alone after the dielectric insertion. What physical principle guarantees that the total charge is conserved?
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CBSE Marking Scheme — 4 Marks (1+1+1+1)
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(i) Capacitance of each capacitor before dielectric insertion:
Formula: C = ε₀A/d
Substituting:
C = (8.854×10⁻¹² × 4×10⁻²) / (2×10⁻³)
C = (3.5416×10⁻¹³) / (2×10⁻³)
C = 1.77×10⁻¹⁰ F
∴ C₁ = C₂ = C ≈ 177 pF
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(ii) Charge and energy stored in series combination (before dielectric):
For two identical capacitors C in series:
C_series = C/2 = 1.77×10⁻¹⁰ / 2 = 0.885×10⁻¹⁰ F = 88.5 pF
Charge stored:
Formula: Q = C_series × V
Q = 0.885×10⁻¹⁰ × 120
∴ Q = 1.062×10⁻⁸ C ≈ 10.6 nC
Energy stored:
Formula: U₀ = ½ C_series V²
U₀ = ½ × 0.885×10⁻¹⁰ × (120)²
U₀ = ½ × 0.885×10⁻¹⁰ × 14400
∴ U₀ = 6.37×10⁻⁷ J ≈ 0.637 μJ
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(iii) New equivalent capacitance after dielectric is inserted into C₁:
When dielectric of constant K = 5 is inserted into C₁:
Formula: C₁' = KC = 5 × 1.77×10⁻¹⁰ = 8.85×10⁻¹⁰ F = 885 pF
C₂ remains unchanged: C₂ = 1.77×10⁻¹⁰ F = 177 pF
New series combination:
Formula: 1/C_new = 1/C₁' + 1/C₂
1/C_new = 1/(8.85×10⁻¹⁰) + 1/(1.77×10⁻¹⁰)
1/C_new = (1.130×10⁹) + (5.650×10⁹)
1/C_new = 6.780×10⁹ F⁻¹
∴ C_new = 1/6.780×10⁹ = 1.475×10⁻¹⁰ F ≈ 147.5 pF
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(iv) New voltage across C₂ after dielectric insertion:
Physical principle: Since the battery was disconnected before the dielectric was inserted, the capacitor combination forms an isolated system. By the principle of conservation of charge (no charge can flow to or from an isolated conductor system), the total charge Q on the series combination remains Q = 1.062×10⁻⁸ C.
In a series combination, the same charge Q resides on each capacitor (charge on each plate is equal and opposite). Therefore, charge on C₂ after insertion = Q = 1.062×10⁻⁸ C.
Voltage across C₂:
Formula: V₂ = Q / C₂
V₂ = (1.062×10⁻⁸) / (1.77×10⁻¹⁰)
∴ V₂ = 60 V
The physical principle is conservation of charge: in an isolated system with no external current path, the net charge on the conductors cannot change. Since the battery is disconnected, no charge can redistribute beyond the capacitor plates, so Q remains constant.
