The Spherical Capacitor Formula gives the capacitance of two concentric spherical conducting shells as \( C = 4\pi\varepsilon_0 \dfrac{ab}{b – a} \), where \( a \) is the inner radius and \( b \) is the outer radius. This formula is a key topic in NCERT Class 12 Physics, Chapter 2 (Electrostatic Potential and Capacitance), and appears regularly in CBSE board exams and JEE Main/Advanced. In this article, you will find the complete derivation, a comprehensive formula sheet, three solved examples at progressive difficulty levels, CBSE exam tips for 2025-26, common mistakes to avoid, and JEE/NEET application strategies.
Key Spherical Capacitor Formulas at a Glance
Quick reference for the most important spherical capacitor formulas.
- Capacitance of spherical capacitor: \( C = 4\pi\varepsilon_0 \dfrac{ab}{b – a} \)
- Isolated spherical conductor: \( C = 4\pi\varepsilon_0 r \)
- Electric field between shells: \( E = \dfrac{Q}{4\pi\varepsilon_0 r^2} \)
- Potential difference: \( V = \dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{1}{a} – \dfrac{1}{b}\right) \)
- Energy stored: \( U = \dfrac{Q^2}{2C} = \dfrac{1}{2}CV^2 \)
- Capacitance with dielectric: \( C = 4\pi\varepsilon_0 K \dfrac{ab}{b – a} \)
- Effective capacitance (series): \( \dfrac{1}{C_{eff}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} \)
What is the Spherical Capacitor Formula?
The Spherical Capacitor Formula defines the capacitance of a system consisting of two concentric hollow spherical conductors. The inner shell carries a charge \( +Q \) and has radius \( a \). The outer shell carries a charge \( -Q \) and has radius \( b \), where \( b > a \). The region between the two shells acts as the dielectric space (usually vacuum or air).
This concept is covered in NCERT Class 12 Physics, Chapter 2 — Electrostatic Potential and Capacitance. The formula is derived using Gauss’s Law and the definition of capacitance \( C = Q/V \). Understanding this formula helps students grasp how geometry influences the ability of a conductor system to store electric charge.
A spherical capacitor is a practical model for understanding capacitance beyond the familiar parallel plate design. It also models an isolated sphere (when the outer shell radius approaches infinity), giving \( C = 4\pi\varepsilon_0 a \). This isolated-sphere result has direct applications in understanding the capacitance of Earth and other large conductors. The Spherical Capacitor Formula is fundamental for both CBSE board preparation and competitive exams like JEE Main, JEE Advanced, and NEET.
Spherical Capacitor Formula — Expression and Variables
The standard expression for the capacitance of a spherical capacitor is:
\[ C = 4\pi\varepsilon_0 \frac{ab}{b – a} \]
where \( a \) is the radius of the inner spherical shell and \( b \) is the radius of the outer spherical shell.
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( C \) | Capacitance of the spherical capacitor | Farad (F) |
| \( \varepsilon_0 \) | Permittivity of free space | C² N¹ m² ( \( 8.85 \times 10^{-12} \) F/m ) |
| \( a \) | Radius of inner spherical shell | Metre (m) |
| \( b \) | Radius of outer spherical shell | Metre (m) |
| \( Q \) | Charge on the inner shell | Coulomb (C) |
| \( V \) | Potential difference between shells | Volt (V) |
| \( K \) | Dielectric constant (relative permittivity) | Dimensionless |
Derivation of the Spherical Capacitor Formula
The derivation uses Gauss’s Law and the definition of capacitance. Follow these steps carefully.
Step 1: Consider a Gaussian surface (a sphere of radius \( r \), where \( a < r < b \)) between the two shells. By Gauss’s Law:
\[ E \cdot 4\pi r^2 = \frac{Q}{\varepsilon_0} \implies E = \frac{Q}{4\pi\varepsilon_0 r^2} \]
Step 2: Calculate the potential difference between the inner shell (radius \( a \)) and the outer shell (radius \( b \)):
\[ V = \int_a^b E \, dr = \frac{Q}{4\pi\varepsilon_0} \int_a^b \frac{dr}{r^2} = \frac{Q}{4\pi\varepsilon_0} \left(\frac{1}{a} – \frac{1}{b}\right) \]
Step 3: Simplify the expression inside the brackets:
\[ V = \frac{Q}{4\pi\varepsilon_0} \cdot \frac{b – a}{ab} \]
Step 4: Apply the definition \( C = Q/V \):
\[ C = \frac{Q}{V} = 4\pi\varepsilon_0 \frac{ab}{b – a} \]
This completes the derivation of the Spherical Capacitor Formula. When \( b \to \infty \), the formula reduces to \( C = 4\pi\varepsilon_0 a \), which is the capacitance of an isolated spherical conductor.
Complete Electrostatics Formula Sheet
The table below covers all key formulas from NCERT Class 12 Chapter 2 that are related to the Spherical Capacitor Formula and capacitance in general.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Spherical Capacitor Capacitance | \( C = 4\pi\varepsilon_0 \dfrac{ab}{b-a} \) | a = inner radius, b = outer radius | Farad (F) | Class 12, Ch 2 |
| Isolated Sphere Capacitance | \( C = 4\pi\varepsilon_0 r \) | r = radius of sphere | Farad (F) | Class 12, Ch 2 |
| Parallel Plate Capacitor | \( C = \dfrac{\varepsilon_0 A}{d} \) | A = area, d = separation | Farad (F) | Class 12, Ch 2 |
| Cylindrical Capacitor | \( C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)} \) | L = length, a = inner radius, b = outer radius | Farad (F) | Class 12, Ch 2 |
| Capacitance with Dielectric | \( C = K \cdot C_0 \) | K = dielectric constant, C₀ = capacitance without dielectric | Farad (F) | Class 12, Ch 2 |
| Energy Stored in Capacitor | \( U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C} \) | C = capacitance, V = voltage, Q = charge | Joule (J) | Class 12, Ch 2 |
| Capacitors in Series | \( \dfrac{1}{C_{eff}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots \) | C₁, C₂ = individual capacitances | Farad (F) | Class 12, Ch 2 |
| Capacitors in Parallel | \( C_{eff} = C_1 + C_2 + \cdots \) | C₁, C₂ = individual capacitances | Farad (F) | Class 12, Ch 2 |
| Electric Field Between Shells | \( E = \dfrac{Q}{4\pi\varepsilon_0 r^2} \) | Q = charge, r = radial distance | N/C or V/m | Class 12, Ch 1 & 2 |
| Potential Difference (Spherical) | \( V = \dfrac{Q}{4\pi\varepsilon_0}\left(\dfrac{1}{a} – \dfrac{1}{b}\right) \) | Q = charge, a = inner radius, b = outer radius | Volt (V) | Class 12, Ch 2 |
| Coulomb’s Law | \( F = \dfrac{1}{4\pi\varepsilon_0} \dfrac{q_1 q_2}{r^2} \) | q₁, q₂ = charges, r = distance | Newton (N) | Class 12, Ch 1 |
| Electric Potential (Point Charge) | \( V = \dfrac{Q}{4\pi\varepsilon_0 r} \) | Q = charge, r = distance | Volt (V) | Class 12, Ch 2 |
Spherical Capacitor Formula — Solved Examples
The following three examples progress from Class 10-level plug-and-calculate to JEE-level concept application. Study each step carefully.
Example 1 (Class 11-12 Level — Direct Application)
Problem: A spherical capacitor has an inner shell of radius 6 cm and an outer shell of radius 9 cm. The space between the shells is air. Calculate the capacitance of the spherical capacitor. (Take \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m.)
Given: \( a = 6 \text{ cm} = 0.06 \text{ m} \), \( b = 9 \text{ cm} = 0.09 \text{ m} \), \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m
Step 1: Write the Spherical Capacitor Formula:
\( C = 4\pi\varepsilon_0 \dfrac{ab}{b – a} \)
Step 2: Calculate \( b – a \):
\( b – a = 0.09 – 0.06 = 0.03 \text{ m} \)
Step 3: Calculate \( ab \):
\( ab = 0.06 \times 0.09 = 5.4 \times 10^{-3} \text{ m}^2 \)
Step 4: Substitute into the formula:
\( C = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{5.4 \times 10^{-3}}{0.03} \)
Step 5: Simplify \( \dfrac{5.4 \times 10^{-3}}{0.03} = 0.18 \text{ m} \)
Step 6: Complete the calculation:
\( C = 4 \times 3.1416 \times 8.85 \times 10^{-12} \times 0.18 \approx 20 \times 10^{-12} \text{ F} = 20 \text{ pF} \)
Answer
Capacitance \( C \approx 20 \) pF
Example 2 (Class 12 / CBSE Board Level — Multi-Step)
Problem: A spherical capacitor with inner radius \( a = 5 \) cm and outer radius \( b = 10 \) cm is filled with a dielectric of dielectric constant \( K = 5 \). A charge of \( Q = 2 \times 10^{-8} \) C is given to the inner shell. Find: (i) the capacitance, and (ii) the energy stored in the capacitor.
Given: \( a = 0.05 \) m, \( b = 0.10 \) m, \( K = 5 \), \( Q = 2 \times 10^{-8} \) C, \( \varepsilon_0 = 8.85 \times 10^{-12} \) F/m
Step 1: Calculate capacitance without dielectric using the Spherical Capacitor Formula:
\( C_0 = 4\pi\varepsilon_0 \dfrac{ab}{b-a} = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{0.05 \times 0.10}{0.10 – 0.05} \)
\( C_0 = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{0.005}{0.05} = 4\pi \times 8.85 \times 10^{-12} \times 0.1 \)
\( C_0 \approx 11.1 \times 10^{-12} \text{ F} = 11.1 \text{ pF} \)
Step 2: Capacitance with dielectric:
\( C = K \cdot C_0 = 5 \times 11.1 \times 10^{-12} = 55.5 \times 10^{-12} \text{ F} \approx 55.5 \text{ pF} \)
Step 3: Calculate energy stored:
\( U = \dfrac{Q^2}{2C} = \dfrac{(2 \times 10^{-8})^2}{2 \times 55.5 \times 10^{-12}} \)
\( U = \dfrac{4 \times 10^{-16}}{1.11 \times 10^{-10}} \approx 3.6 \times 10^{-6} \text{ J} = 3.6 \text{ \( \mu \)J} \)
Answer
(i) Capacitance \( C \approx 55.5 \) pF
(ii) Energy stored \( U \approx 3.6 \) µJ
Example 3 (JEE/NEET Level — Concept Application)
Problem: Two spherical capacitors are connected in series. The first has inner radius 3 cm and outer radius 6 cm. The second has inner radius 4 cm and outer radius 8 cm. Both are filled with air. Find the effective capacitance of the combination.
Given: For capacitor 1: \( a_1 = 0.03 \) m, \( b_1 = 0.06 \) m. For capacitor 2: \( a_2 = 0.04 \) m, \( b_2 = 0.08 \) m.
Step 1: Calculate \( C_1 \) using the Spherical Capacitor Formula:
\( C_1 = 4\pi\varepsilon_0 \dfrac{a_1 b_1}{b_1 – a_1} = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{0.03 \times 0.06}{0.03} \)
\( C_1 = 4\pi \times 8.85 \times 10^{-12} \times 0.06 \approx 6.67 \times 10^{-12} \text{ F} = 6.67 \text{ pF} \)
Step 2: Calculate \( C_2 \):
\( C_2 = 4\pi\varepsilon_0 \dfrac{a_2 b_2}{b_2 – a_2} = 4\pi \times 8.85 \times 10^{-12} \times \dfrac{0.04 \times 0.08}{0.04} \)
\( C_2 = 4\pi \times 8.85 \times 10^{-12} \times 0.08 \approx 8.90 \times 10^{-12} \text{ F} = 8.90 \text{ pF} \)
Step 3: Apply the series combination formula:
\( \dfrac{1}{C_{eff}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{1}{6.67} + \dfrac{1}{8.90} \) (all in pF)
\( \dfrac{1}{C_{eff}} = 0.1499 + 0.1124 = 0.2623 \text{ pF}^{-1} \)
Step 4: Find \( C_{eff} \):
\( C_{eff} = \dfrac{1}{0.2623} \approx 3.81 \text{ pF} \)
Answer
Effective capacitance \( C_{eff} \approx 3.81 \) pF
CBSE Exam Tips 2025-26
- Always convert radii to metres before substituting into the Spherical Capacitor Formula. A very common error is using centimetres directly, which gives an answer off by a factor of 100.
- Memorise the value of \( 4\pi\varepsilon_0 \): \( 4\pi\varepsilon_0 \approx \dfrac{1}{9 \times 10^9} \) F/m. This shortcut saves calculation time in board exams.
- Know the limiting case: When \( b \to \infty \), the spherical capacitor formula reduces to \( C = 4\pi\varepsilon_0 a \). CBSE has asked this limiting case as a short-answer question in previous years.
- Show derivation steps clearly. The CBSE 5-mark derivation question for spherical capacitors requires you to show the Gauss’s Law step, the integration step, and the final \( C = Q/V \) step. We recommend writing each step on a separate line.
- Practice energy stored problems. CBSE 2024 and 2023 papers both included a capacitor energy question. Use \( U = Q^2/(2C) \) when charge is given and \( U = \frac{1}{2}CV^2 \) when voltage is given.
- Revise dielectric insertion. When a dielectric of constant \( K \) fills the space, multiply \( C \) by \( K \). This is a frequent 2-mark question in CBSE 2025-26 pattern papers.
Common Mistakes to Avoid
Students frequently lose marks on the Spherical Capacitor Formula due to avoidable errors. Here are the most common mistakes and how to correct them.
- Mistake 1 — Swapping \( a \) and \( b \): Always remember \( a \) is the inner (smaller) radius and \( b \) is the outer (larger) radius. Swapping them gives a negative denominator and a physically meaningless negative capacitance.
- Mistake 2 — Using \( b – a \) in the numerator: The numerator is the product \( ab \), not the difference \( b – a \). The difference goes in the denominator. Write the formula carefully in every step.
- Mistake 3 — Forgetting unit conversion: Radii given in centimetres must be converted to metres before substitution. Capacitance in Farads is the standard SI unit expected in board and JEE answers.
- Mistake 4 — Ignoring the dielectric constant: When the problem states that a dielectric fills the gap, multiply the air-capacitance result by \( K \). Many students forget this step and lose the full marks for that part.
- Mistake 5 — Confusing series and parallel for capacitors: In a series combination, the reciprocals of capacitances add. In a parallel combination, the capacitances add directly. This is opposite to the rule for resistors, which confuses many students.
JEE/NEET Application of Spherical Capacitor Formula
In our experience, JEE aspirants encounter the Spherical Capacitor Formula in two to three questions per year across JEE Main and Advanced papers. NEET aspirants see it less frequently, but it appears in the context of energy stored and dielectric insertion. Here are the key application patterns to master.
Pattern 1: Finding Capacitance and Comparing Geometries
JEE problems often ask you to compare the capacitance of a spherical capacitor with a parallel plate capacitor of equivalent dimensions. The key insight is that capacitance depends on geometry. For a spherical capacitor, \( C = 4\pi\varepsilon_0 ab/(b-a) \). For a parallel plate capacitor, \( C = \varepsilon_0 A/d \). JEE questions may ask which geometry gives higher capacitance for a given volume. Practise both formulas together.
Pattern 2: Energy and Charge Distribution
JEE Advanced problems frequently combine the Spherical Capacitor Formula with energy methods. A typical question gives the potential difference and asks for the charge, energy, and electric field at a specific radius. Use \( Q = CV \), \( U = Q^2/(2C) \), and \( E = Q/(4\pi\varepsilon_0 r^2) \) in sequence. In our experience, writing all three formulas at the top of your solution saves time and earns method marks.
Pattern 3: Earthing One Shell
A classic JEE problem involves earthing the outer shell of a spherical capacitor. When the outer shell is earthed (connected to ground), its potential becomes zero. The capacitance remains \( C = 4\pi\varepsilon_0 ab/(b-a) \). However, if the inner shell is earthed and the outer shell is given a charge, the effective capacitance changes. This conceptual twist is a favourite in JEE Advanced. Our experts suggest drawing the configuration before writing any formula.
For NEET, focus on the definition, the limiting case of an isolated sphere, and energy stored calculations. These three topics cover over 90% of NEET questions on this topic. Visit ncert.nic.in to access the official NCERT Class 12 Physics textbook for the standard derivation and examples.
FAQs on Spherical Capacitor Formula
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