Question

Let $E_1\left(r\right),E_2\left(r\right)$ and $E_3\left(r\right)$ be the respective electric fields at a distance $r$ from a point charge $Q$, an infinitely long wire with constant linear charge density $\lambda$, an infinite plane with uniform surface charge density $\sigma$.
If $E_1\left(r_0\right)=E_2\left(r_0\right)=E_3\left(r_0\right)$ at a given distance $r_0$, then

• A. $Q=4\sigma \pi r_0^2$
• B. $r_0=\frac{\lambda }{2\pi \sigma }$
• C. $Q=2\sigma \pi r_0^2$
• D. $r_0=\frac{\lambda }{\pi \sigma }$

Answer 1

Answer: (C), (D)

The correct options are
C $Q=2\sigma \pi r_0^2$
D $r_0=\frac{\lambda }{\pi \sigma }$


Electric field at a distance $r$ from a point charge $Q$ is given by
$E=\frac{kQ}{r^2}$
$\therefore E_1=\frac{kQ}{r_0^2}$


Electric field at a distance $r$ from an infinitely long wire of linear chage density $\lambda$ is given by
$E=\frac{\lambda }{2\pi {ϵ}_{0}r}$
$\therefore E_2=\frac{\lambda }{2\pi {ϵ}_0{r}_0}$


Electric field at a distance $r$ from an infinite plane of surface chage density $\sigma$ is given by
$E=\frac{\sigma }{2{ϵ}_0}$
$\therefore E_3=\frac{\sigma }{2{ϵ}_0}$

Given $E_1=E_2=E_3$
$E_1=E_3\Rightarrow \frac{kQ}{r_0^2}=\frac{\sigma }{2{ϵ}_0}$
$\Rightarrow Q=\frac{r_o^2}{k}\times \frac{\sigma }{2{ϵ}_0}$
$\Rightarrow Q=r_o^2\times 4\pi {ϵ}_0\times \frac{\sigma }{2{ϵ}_0}$ ($\because k=\frac{1}{4\pi {ϵ}_0}$)
$\Rightarrow Q=2\pi \sigma r_0^2$ (c)

Also, $E_2=E_3\Rightarrow \frac{\lambda }{2\pi {ϵ}_0{r}_0}=\frac{\sigma }{2{ϵ}_0}$
$\Rightarrow r_0=\frac{\lambda }{\pi \sigma }$ (d)