Subject: Applied Physics 2

Topic: Electrodynamics

Difficulty: Medium

$$ Formula: r= √(x^2+y^2+z^2 ) $$ $$ ∅=〖tan〗^(-1) y/x $$ $$ ∅= 〖tan〗^(-1) √(x^2+y^2 )/z $$

$ \text{ Solution: Given x = 2√3 , y = 6 z = -4 } \\ $

$ r= √(x^2+y^2+z^2 ) \\ $

$ \text{= √(〖2√3〗^2+6^2+〖-4〗^2 )} \\ $

$ = 8 \\ $

$ tan∅= y/x = 6/(2√3)= √3 \\ $

$ \text { Therefore φ = π/3 } \\$ $ tanθ= √(x^2+y^2 )/z= √(〖2√3〗^2+6^2 )/(-4)= (4√3)/(-4)= -√3 \ $ $$ Θ = 2π /3 $$