The y = 0 plane separates air in the upper half-space (y > 0) from a good conductor (y ≤ 0) while a surface charge density, Ps= C₁ sin ßx, and a surface current density, K = C₂ cos ßx, are present at the interface. (C₁,C₂ and 3 are constants). 1 Find expressions for E and H in the air at the interface.
The y = 0 plane separates air in the upper half-space (y > 0) from a good conductor (y ≤ 0) while a surface charge density, Ps= C₁ sin ßx, and a surface current density, K = C₂ cos ßx, are present at the interface. (C₁,C₂ and 3 are constants). 1 Find expressions for E and H in the air at the interface.
Related questions
Question

Transcribed Image Text:Exercise 10.
The y = 0 plane separates air in the upper half-space (y > 0) from a good conductor (y ≤ 0) while a
surface charge density, p = C₁ sin ßx, and a surface current density, K = C₂ cos ßx, are present at
the interface. (C₁,C₂ and ß are constants).
Find expressions for E and H in the air at the interface.
Ans.
E = 9¹sin ßx; H = 2C₂ cos Bx
3
Expert Solution

This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 3 steps
