Let E_1(r),E_2(r),and E_3(r) be the respectiv | vedclass

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(C) The electric fields are given by: $E_1(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$,$E_2(r) = \frac{\lambda}{2 \pi \epsilon_0 r}$,and $E_3(r) = \frac{\sig

Vedclass · Accepted Answer

$E_1(r_0 / 2) = 2 E_2(r_0 / 2)$

Vedclass · Answer

(C) The electric fields are given by: $E_1(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$,$E_2(r) = \frac{\lambda}{2 \pi \epsilon_0 r}$,and $E_3(r) = \frac{\sigma}{2 \epsilon_0}$.Given $E_1(r_0) = E_2(r_0) = E_3(r_0) = E_0$,we have:$\frac{Q}{4 \pi \epsilon_0 r_0^2} = \frac{\lambda}{2 \pi \epsilon_0 r_0} = \frac{\sigma}{2 \epsilon_0} = E_0$.From $E_2(r_0) = E_3(r_0)$,we get $\frac{\lambda}{2 \pi \epsilon_0 r_0} = \frac{\sigma}{2 \epsilon_0}$,which implies $r_0 = \frac{\lambda}{\pi \sigma}$. Thus,option $B$ is incorrect.From $E_1(r_0) = E_3(r_0)$,we get $\frac{Q}{4 \pi \epsilon_0 r_0^2} = \frac{\sigma}{2 \epsilon_0}$,which implies $Q = 2 \pi \sigma r_0^2$. Thus,option $A$ is incorrect.Now,check option $C$: $E_1(r_0/2) = \frac{Q}{4 \pi \epsilon_0 (r_0/2)^2} = 4 E_1(r_0) = 4 E_0$. Also,$E_2(r_0/2) = \frac{\lambda}{2 \pi \epsilon_0 (r_0/2)} = 2 E_2(r_0) = 2 E_0$. Therefore,$E_1(r_0/2) = 2 E_2(r_0/2)$. Option $C$ is correct.Check option $D$: $E_3(r)$ is independent of $r$,so $E_3(r_0/2) = E_3(r_0) = E_0$. Since $E_2(r_0/2) = 2 E_0$,$E_2(r_0/2) = 2 E_3(r_0/2)$. Option $D$ is incorrect.

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