Question

Infinite charge sheet in xy plane of surface charge density $\sigma$ and infinite long wire of linear charge density $\lambda$ placed at $(0,0,4)$ and $\sigma=2 \lambda$. Then net electric field $(0,0,2)$.

• A. $Q=4\sigma \pi r^2_0$
• B. $r _{0}=\frac{\lambda}{2 \pi \sigma}$
• C. $E _{1}\left( r _{0} / 2\right)=2 E _{2}\left( r _{0} / 2\right)$
• D. $E _{2}\left( r _{0} / 2\right)=4 E _{3}\left( r _{0} / 2\right)$

Answer 1

Answer: (C)

$E _{1}\left( r _{0} / 2\right)=2 E _{2}\left( r _{0} / 2\right)$

Answer 2

$\frac{ Q }{4 \pi \varepsilon_{0} r _{0}^{2}}=\frac{\lambda}{2 \pi \varepsilon_{0} I _{0}}=\frac{\sigma}{2 \varepsilon_{0}}$
$E _{1}\left(\frac{ r _{0}}{2}\right)=\frac{ Q }{\pi \varepsilon_{0} I _{0}^{2}},$
$E _{2}\left(\frac{ r _{0}}{2}\right)=\frac{\lambda}{\pi \varepsilon_{0} I _{0}},$
$E _{3}\left(\frac{ r _{0}}{2}\right)=\frac{\sigma}{2 \varepsilon_{0}}$
$\therefore E _{1}\left(\frac{ r _{0}}{2}\right)$
$=2 E _{2}\left(\frac{ r _{0}}{2}\right)$