Mumbai University > Electronics Engineering > Sem 5 > Electromagnetic Engineering

Marks: 10 Marks

Year: May 2015

For practical case we will analyze this given can figuration in cylindrical co-ordinate system r,∅,z

Charge for $\dfrac{dL}{dz}$ section is given as

$dQ=ρ_L .d_L=ρ_L.d_z$

As $d_z- 0$ , we can apply point charge concept for $d_z$ section.

$d \bar{E}=\dfrac{dQ}{4πεR^2} \overline{a_R} \\ d \bar{E}=\dfrac{ρ_L.d_z}{4πεR^2} \overline{a_R} \\ \bar{E}= ∫_{z= -∞}^∞ \dfrac{ρ_L.d_z}{4πεR^2} \overline{a_R}………(1) \\ Where, \\ \overline{a_R}=\dfrac{r. \overline{a_R}- z. \overline{a_z}}{\sqrt{(r^2+ z^2 )}} ≈\dfrac{r.\overline{a_R}}{\sqrt{r^2+ z^2}}$

Substituting in equation (1)

$\bar{E}=∫_{z= -∞}^∞\dfrac{ρ_L.d_z.r.\overline{a_R}}{4πε (r^2+ z^2 )}^{\frac{3}{2}} \\ \bar{E}=\dfrac{ρ_L.r.\overline{a_R}}{4πε} ∫_{-∞}^∞ \dfrac{d_z}{r^2+ z^2}^\frac{3}{2} \\ Put z=r. \tanθ \\ d_z=r. \sec^2⁡θ .dθ \\ \bar{E}=\dfrac{ρ_L.r.\overline{a_R}}{4πε} ∫_{\frac{–π}{2}}^\frac{π}{2}\dfrac{r. \sec^2⁡θ .dθ}{r^2+r^2 \tan^2⁡θ }^\frac{3}{2} \\ \bar{E}=\dfrac{ρ_L.r^2.\overline{a_R}}{4πε r^3} ∫_\frac{-π}{2}^\frac{π}{2}dθ/ \secθ \\ \bar{E}=\dfrac{ρ_L.\overline{a_R}}{4πεr} [\sinθ]_\frac{-π}{2}^\frac{π}{2} \\ \bar{E}=\dfrac{ρ_L.\overline{a_R}}{2π ε_r }…….v/m$