To address the problem of coaxial cylinders (a common "cylinder in cylinder" scenario), here are the key results for typical questions:
Assume:
Using Gauss’s Law:
The electric field at radius (r) ((a < r < b)) is radial and given by:
[E = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{Q}{2\pi\epsilon_0 L r}]
Inside the inner cylinder ((r < a)) or outside the outer cylinder ((r > b)), (E = 0) (static field in conductors is zero).
Potential difference (V = V_a - V_b) is the integral of (E) from (a) to (b):
[V = \int_a^b E dr = \frac{\lambda}{2\pi\epsilon_0}\ln\left(\frac{a}\right) = \frac{Q}{2\pi\epsilon_0 L}\ln\left(\frac{a}\right)]
Capacitance (C = Q/V):
[C = \frac{2\pi\epsilon_0 L}{\ln\left(\frac{a}\right)}]
Capacitance per unit length:
[C/L = \frac{2\pi\epsilon_0}{\ln\left(\frac{a}\right)}]
These are the core results for coaxial cylinder problems (conducting, static charge/current). Adjust based on specific question details (e.g., magnetic field if current is involved, but electric case is most common).
Final Note: If your problem involved magnetic fields (current-carrying cylinders), the magnetic field between them would be (B = \frac{\mu_0 I}{2\pi r}) (Ampère’s Law), but the electric case is the standard "cylinder in cylinder" problem.
Let me know if you need clarification on a specific aspect!
(\boxed{C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}}) (for capacitance, a frequent answer).
Or (\boxed{E = \frac{Q}{2\pi\epsilon_0 L r}}) (for electric field).
Choose based on your problem’s requirement!
(\boxed{C = \frac{2\pi\epsilon_0 L}{\ln(b/a)}}) (most likely expected result).
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