The Cylindrical Capacitor Formula gives the capacitance of a capacitor formed by two coaxial cylindrical conductors as C = 2πε₀L / ln(b/a), where L is the length of the cylinder, b is the outer radius, and a is the inner radius. This formula is a key concept in Class 12 Physics (NCERT Chapter 2 — Electrostatic Potential and Capacitance) and appears regularly in JEE Main, JEE Advanced, and NEET examinations. In this article, we cover the complete derivation, a formula cheat sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET application patterns.
Key Cylindrical Capacitor Formulas at a Glance
Quick reference for the most important cylindrical capacitor formulas.
- Capacitance of cylindrical capacitor: \( C = \dfrac{2\pi\varepsilon_0 L}{\ln(b/a)} \)
- With dielectric medium: \( C = \dfrac{2\pi\varepsilon_0 \varepsilon_r L}{\ln(b/a)} \)
- Electric field between cylinders: \( E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \)
- Potential difference: \( V = \dfrac{\lambda}{2\pi\varepsilon_0} \ln\left(\dfrac{b}{a}\right) \)
- Charge stored: \( Q = CV \)
- Energy stored: \( U = \dfrac{Q^2}{2C} = \dfrac{1}{2}CV^2 \)
- Capacitance per unit length: \( \dfrac{C}{L} = \dfrac{2\pi\varepsilon_0}{\ln(b/a)} \)
What is the Cylindrical Capacitor Formula?
The Cylindrical Capacitor Formula describes the capacitance of a system consisting of two long, coaxial cylindrical conductors separated by a dielectric or vacuum. The inner cylinder has radius a and the outer cylinder has radius b, with b > a. The length of the system is L.
This concept is introduced in NCERT Class 12 Physics, Chapter 2 — Electrostatic Potential and Capacitance. It extends the idea of a parallel plate capacitor to a cylindrical geometry. The formula is derived using Gauss's Law applied to the electric field between the two cylinders.
A cylindrical capacitor is not just a theoretical construct. It has real-world applications in coaxial cables, cylindrical sensors, and certain types of electronic components. Understanding this formula helps students master the broader topic of capacitors, which carries significant weight in both CBSE board exams and competitive entrance tests like JEE and NEET.
The capacitance depends on the ratio of the radii, not their individual values. This is a crucial insight. Doubling both radii while keeping their ratio constant leaves the capacitance unchanged.
Cylindrical Capacitor Formula — Expression and Variables
The standard expression for the capacitance of a cylindrical capacitor is:
\[ C = \frac{2\pi\varepsilon_0 L}{\ln\left(\frac{b}{a}\right)} \]
When the space between the cylinders is filled with a dielectric of relative permittivity \( \varepsilon_r \), the formula becomes:
\[ C = \frac{2\pi\varepsilon_0 \varepsilon_r L}{\ln\left(\frac{b}{a}\right)} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( C \) | Capacitance of the cylindrical capacitor | Farad (F) |
| \( \varepsilon_0 \) | Permittivity of free space (8.854 × 10⁻¹² F/m) | F m⁻¹ |
| \( \varepsilon_r \) | Relative permittivity (dielectric constant) | Dimensionless |
| \( L \) | Length of the cylindrical capacitor | Metre (m) |
| \( a \) | Radius of the inner cylinder | Metre (m) |
| \( b \) | Radius of the outer cylinder | Metre (m) |
| \( \ln(b/a) \) | Natural logarithm of the ratio of radii | Dimensionless |
Derivation of the Cylindrical Capacitor Formula
We derive the formula using Gauss's Law. Consider a cylindrical Gaussian surface of radius r (where a < r < b) and length L coaxial with the capacitor.
Step 1: Apply Gauss's Law. The electric flux through the Gaussian surface equals the enclosed charge divided by \( \varepsilon_0 \):
\[ E \cdot 2\pi r L = \frac{Q}{\varepsilon_0} \]
Step 2: Solve for the electric field:
\[ E = \frac{Q}{2\pi\varepsilon_0 L r} = \frac{\lambda}{2\pi\varepsilon_0 r} \]
where \( \lambda = Q/L \) is the linear charge density.
Step 3: Find the potential difference between the cylinders by integrating the electric field from a to b:
\[ V = \int_a^b E \, dr = \frac{Q}{2\pi\varepsilon_0 L} \int_a^b \frac{dr}{r} = \frac{Q}{2\pi\varepsilon_0 L} \ln\left(\frac{b}{a}\right) \]
Step 4: Use the definition of capacitance \( C = Q/V \):
\[ C = \frac{Q}{V} = \frac{2\pi\varepsilon_0 L}{\ln(b/a)} \]
This completes the derivation of the Cylindrical Capacitor Formula.
Complete Electrostatics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| 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 |
| Parallel Plate Capacitor | \( C = \dfrac{\varepsilon_0 A}{d} \) | A = plate area, d = separation | Farad (F) | Class 12, Ch 2 |
| Spherical Capacitor | \( C = 4\pi\varepsilon_0 \dfrac{ab}{b-a} \) | a = inner radius, b = outer radius | Farad (F) | Class 12, Ch 2 |
| Capacitors in Series | \( \dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots \) | C₁, C₂ = individual capacitances | Farad (F) | Class 12, Ch 2 |
| Capacitors in Parallel | \( C_{eq} = C_1 + C_2 + \cdots \) | C₁, C₂ = individual capacitances | 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 |
| Electric Field (Cylindrical) | \( E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \) | λ = linear charge density, r = radial distance | V m⁻¹ | Class 12, Ch 1 |
| Capacitance with Dielectric | \( C = \varepsilon_r C_0 \) | εᵣ = dielectric constant, C₀ = vacuum capacitance | Farad (F) | Class 12, Ch 2 |
| Charge on Capacitor | \( Q = CV \) | C = capacitance, V = potential difference | Coulomb (C) | Class 12, Ch 2 |
| Capacitance per Unit Length (Cylindrical) | \( \dfrac{C}{L} = \dfrac{2\pi\varepsilon_0}{\ln(b/a)} \) | a = inner radius, b = outer radius | F m⁻¹ | Class 12, Ch 2 |
Cylindrical Capacitor Formula — Solved Examples
Example 1 (Class 11-12 Level — Direct Application)
Problem: A cylindrical capacitor has an inner radius of 1 cm, an outer radius of 3 cm, and a length of 20 cm. Calculate its capacitance. (Take \( \varepsilon_0 = 8.854 \times 10^{-12} \) F/m)
Given:
- \( a = 1 \) cm \( = 0.01 \) m
- \( b = 3 \) cm \( = 0.03 \) m
- \( L = 20 \) cm \( = 0.20 \) m
- \( \varepsilon_0 = 8.854 \times 10^{-12} \) F/m
Step 1: Write the Cylindrical Capacitor Formula:
\[ C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)} \]
Step 2: Calculate the ratio \( b/a \):
\( b/a = 0.03 / 0.01 = 3 \)
Step 3: Find \( \ln(3) \approx 1.0986 \)
Step 4: Substitute all values:
\[ C = \frac{2 \times 3.1416 \times 8.854 \times 10^{-12} \times 0.20}{1.0986} \]
\[ C = \frac{2 \times 3.1416 \times 1.7708 \times 10^{-12}}{1.0986} \]
\[ C = \frac{11.123 \times 10^{-12}}{1.0986} \approx 10.13 \times 10^{-12} \text{ F} \]
Answer
Capacitance \( C \approx 10.13 \) pF
Example 2 (Class 12 Level — With Dielectric)
Problem: The space between the cylinders of a cylindrical capacitor (inner radius = 2 mm, outer radius = 6 mm, length = 50 cm) is completely filled with a dielectric of relative permittivity \( \varepsilon_r = 4 \). Find the capacitance and the charge stored when a potential difference of 100 V is applied.
Given:
- \( a = 2 \) mm \( = 2 \times 10^{-3} \) m
- \( b = 6 \) mm \( = 6 \times 10^{-3} \) m
- \( L = 0.50 \) m
- \( \varepsilon_r = 4 \)
- \( V = 100 \) V
Step 1: Use the formula with dielectric:
\[ C = \frac{2\pi\varepsilon_0 \varepsilon_r L}{\ln(b/a)} \]
Step 2: Calculate \( \ln(b/a) = \ln(6/2) = \ln(3) \approx 1.0986 \)
Step 3: Substitute values:
\[ C = \frac{2\pi \times 8.854 \times 10^{-12} \times 4 \times 0.50}{1.0986} \]
\[ C = \frac{2 \times 3.1416 \times 8.854 \times 10^{-12} \times 2}{1.0986} \]
\[ C = \frac{111.26 \times 10^{-12}}{1.0986} \approx 101.3 \times 10^{-12} \text{ F} = 101.3 \text{ pF} \]
Step 4: Find the charge stored using \( Q = CV \):
\[ Q = 101.3 \times 10^{-12} \times 100 = 1.013 \times 10^{-8} \text{ C} \approx 10.13 \text{ nC} \]
Answer
Capacitance \( C \approx 101.3 \) pF; Charge stored \( Q \approx 10.13 \) nC
Example 3 (JEE/NEET Level — Concept Application)
Problem: A cylindrical capacitor of length 1 m has inner and outer radii in the ratio 1 : e (where e ≈ 2.718 is Euler's number). It is connected to a 200 V battery. Find: (a) the capacitance, (b) the energy stored, and (c) the electric field at a point midway between the two cylinders if the inner radius is 5 cm.
Given:
- \( L = 1 \) m, \( V = 200 \) V
- \( b/a = e \), so \( \ln(b/a) = \ln(e) = 1 \)
- \( a = 5 \) cm \( = 0.05 \) m, so \( b = 0.05e \approx 0.1359 \) m
Step 1 — Capacitance:
\[ C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)} = \frac{2\pi \times 8.854 \times 10^{-12} \times 1}{1} = 55.63 \times 10^{-12} \text{ F} \approx 55.6 \text{ pF} \]
Step 2 — Energy stored:
\[ U = \frac{1}{2}CV^2 = \frac{1}{2} \times 55.63 \times 10^{-12} \times (200)^2 \]
\[ U = \frac{1}{2} \times 55.63 \times 10^{-12} \times 40000 = 1.113 \times 10^{-6} \text{ J} \approx 1.11 \text{ μJ} \]
Step 3 — Charge on capacitor:
\[ Q = CV = 55.63 \times 10^{-12} \times 200 = 11.126 \times 10^{-9} \text{ C} \]
Step 4 — Midpoint radius: The midpoint between the two cylinders is at \( r = (a + b)/2 = (0.05 + 0.1359)/2 \approx 0.0930 \) m.
Step 5 — Electric field at midpoint:
\[ E = \frac{Q}{2\pi\varepsilon_0 L r} = \frac{11.126 \times 10^{-9}}{2\pi \times 8.854 \times 10^{-12} \times 1 \times 0.0930} \]
\[ E = \frac{11.126 \times 10^{-9}}{5.173 \times 10^{-12}} \approx 2150 \text{ V/m} \]
Answer
(a) \( C \approx 55.6 \) pF, (b) \( U \approx 1.11 \) μJ, (c) \( E \approx 2150 \) V/m
CBSE Exam Tips 2025-26
- Memorise the formula structure: The Cylindrical Capacitor Formula \( C = 2\pi\varepsilon_0 L / \ln(b/a) \) is directly asked in 3-mark and 5-mark questions. Write it clearly in exams.
- Show derivation steps: CBSE awards marks for each step of the Gauss's Law derivation. Never skip intermediate steps, especially the integration step.
- Convert units carefully: Radii are often given in cm or mm. Always convert to metres before substituting. We recommend writing unit conversions explicitly.
- Remember \( \ln \) not \( \log_{10} \): The formula uses the natural logarithm (base e). Using log base 10 is a very common error that costs full marks.
- Dielectric modification: If a dielectric is mentioned, multiply \( \varepsilon_0 \) by \( \varepsilon_r \). This is a frequent twist in 2025-26 paper patterns.
- Link to coaxial cables: CBSE questions sometimes ask for the real-life application. A coaxial cable is the most important practical example of a cylindrical capacitor.
Common Mistakes to Avoid
- Using log₁₀ instead of ln: The natural logarithm (ln) must be used in the Cylindrical Capacitor Formula. Using log₁₀ gives a result that is approximately 2.303 times too small. Always verify which logarithm base is required.
- Swapping a and b: The inner radius is always a and the outer radius is always b. Since b > a, we have ln(b/a) > 0. Swapping them gives a negative logarithm, which is physically meaningless.
- Forgetting to convert units: Radii given in centimetres or millimetres must be converted to metres. Capacitance values in pF or nF must be converted to Farads for energy calculations.
- Omitting the length L: Capacitance is proportional to length. Some students use the formula for capacitance per unit length and forget to multiply by L. Always check whether L is included.
- Ignoring the dielectric constant: When a dielectric fills the space between the cylinders, the capacitance increases by a factor of \( \varepsilon_r \). Treating the medium as vacuum when a dielectric is present is a common error in JEE problems.
JEE/NEET Application of Cylindrical Capacitor Formula
In our experience, JEE aspirants encounter the Cylindrical Capacitor Formula in several distinct question patterns. Understanding these patterns builds both speed and accuracy in the exam hall.
Pattern 1 — Capacitance Calculation with Given Geometry
JEE Main frequently asks students to calculate the capacitance of a cylindrical capacitor with given radii and length. The key skill is computing \( \ln(b/a) \) quickly. Memorising \( \ln 2 \approx 0.693 \), \( \ln 3 \approx 1.099 \), and \( \ln e = 1 \) saves valuable time. Problems often use ratios like \( b/a = e \) specifically so that \( \ln(b/a) = 1 \), simplifying the calculation.
Pattern 2 — Energy and Charge Storage
JEE Advanced problems combine the Cylindrical Capacitor Formula with energy storage concepts. A typical question asks for the energy stored after connecting to a battery, or the heat dissipated when two capacitors are connected. The approach is: (i) find C using the cylindrical formula, (ii) find Q = CV, and (iii) apply \( U = Q^2/(2C) \) or \( U = CV^2/2 \). NEET questions at this level are more straightforward but test the same concept.
Pattern 3 — Electric Field Between Cylinders
JEE Advanced tests the electric field expression \( E = \lambda / (2\pi\varepsilon_0 r) \) derived during the capacitor derivation. Students must know that the electric field is non-uniform between the cylinders — it varies as 1/r. This contrasts with the uniform field in a parallel plate capacitor. Questions may ask for the field at a specific radius, or for the ratio of fields at two different radii. Our experts suggest practising both the field expression and its integration to potential difference.
For NEET, the Cylindrical Capacitor Formula itself is less frequently tested as a direct calculation. However, conceptual questions about the effect of inserting a dielectric, changing the length, or doubling the radii appear regularly. Knowing that capacitance scales with L and inversely with ln(b/a) is sufficient for most NEET questions.
For further reading on related electrostatics topics, visit the official NCERT Class 12 Physics Chapter 2 on ncert.nic.in.
FAQs on Cylindrical Capacitor Formula
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