"Current Between Coaxial Cylinders: Griffiths Electrodynamics Example 7.2 Solved !

In this video, we solve Example 7.2 from David J. Griffiths' Introduction to Electrodynamics, where we calculate the current flowing between two long coaxial metal cylinders (radii 𝑎 and 𝑏) separated by a material of conductivity σ and maintained at a potential difference 𝑉. What You’ll Learn: ✔ Derive the current 𝑰 flowing from one cylinder to the other over length 𝐿. ✔ Apply Ohm’s law in cylindrical coordinates for conducting materials. ✔ Relate current density (𝑱), electric field (𝑬), and conductivity (σ). ✔ Understand boundary conditions for potentials in cylindrical systems. Key Steps: Use Gauss’s law to find the electric field 𝑬 between the cylinders. Relate 𝑬 to current density 𝑱 via 𝑱 = σ𝑬. Integrate 𝑱 over the cylindrical surface to find total current 𝑰. Formula Derived: I = 2 π L σ V ln ⁡ ( b / a ) I= ln(b/a) 2πLσV ​ Textbook Reference: Introduction to Electrodynamics, 4th Edition (Example 7.2) Who Is This For? Students studying electrodynamics (Griffiths or similar textbooks). Anyone preparing for physics exams or interested in applied electromagnetism. Timestamps: 00:00 - Problem Statement 01:00 - Electric Field in Cylindrical Coordinates 03:00 - linear charge Density 05:30 - Calculating Total Current Call to Action: 👍 Like if this helped! 🔔 Subscribe for more Griffiths problem walkthroughs. 💬 Comment below with requests for other problems or chapters!