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(NONE) For configuration $I$: The electric field in any region between the sheets is given by $E = \frac{1}{2\epsilon_0} \sum \sigma_i$. In region $4$

Vedclass · Accepted Answer

In region $4$ of the configuration $I$,the magnitude of the electric field is zero.

Vedclass · Answer

(NONE) For configuration $I$: The electric field in any region between the sheets is given by $E = \frac{1}{2\epsilon_0} \sum \sigma_i$. In region $4$,$E_4 = \frac{1}{2\epsilon_0} (\sigma_0 - \sigma_0 + \sigma_0 - \sigma_0 + \sigma_0) = \frac{\sigma_0}{2\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0}{2\epsilon_0}$. Thus,option $C$ is incorrect. The potential difference between the first and last sheet is $V_1 - V_6 = E_1 d + E_2 d + E_3 d + E_4 d + E_5 d$. For configuration $I$,$E_1 = \frac{\sigma_0}{2\epsilon_0}$,$E_2 = -\frac{\sigma_0}{2\epsilon_0}$,$E_3 = \frac{\sigma_0}{2\epsilon_0}$,$E_4 = -\frac{\sigma_0}{2\epsilon_0}$,$E_5 = \frac{\sigma_0}{2\epsilon_0}$. $V_1 - V_6 = d \frac{\sigma_0}{2\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0 d}{2\epsilon_0} = \frac{9 	imes 10^{-6} 	imes 10^{-6}}{2 	imes 9 	imes 10^{-12}} = 0.5 V$. Option $B$ is incorrect. For configuration $II$: The electric field in region $3$ is $E_3 = \frac{1}{2\epsilon_0} (\frac{\sigma_0}{2} - \sigma_0 + \sigma_0) = \frac{\sigma_0}{4\epsilon_0}$. Option $A$ is incorrect. The potential difference between the first and last sheet for configuration $II$: $E_1 = \frac{\sigma_0}{4\epsilon_0}$,$E_2 = -\frac{\sigma_0}{4\epsilon_0}$,$E_3 = \frac{\sigma_0}{4\epsilon_0}$,$E_4 = -\frac{\sigma_0}{4\epsilon_0}$,$E_5 = \frac{\sigma_0}{4\epsilon_0}$. $V_1 - V_6 = d (E_1 + E_2 + E_3 + E_4 + E_5) = d \frac{\sigma_0}{4\epsilon_0} (1 - 1 + 1 - 1 + 1) = \frac{\sigma_0 d}{4\epsilon_0} 
eq 0$. Option $D$ is incorrect. Upon re-evaluating the configurations,no provided option is correct based on standard physics principles.

List-$I$	List-$II$
$P$. $E$ is independent of $d$	$1$. $A$ point charge $Q$ at the origin
$Q$. $E \propto \frac{1}{d}$	$2$. $A$ small dipole with point charges $Q$ at $(0, 0, l)$ and $-Q$ at $(0, 0, -l)$. Take $2l \ll d$.
$R$. $E \propto \frac{1}{d^2}$	$3$. An infinite line charge coincident with the $x$-axis,with uniform linear charge density $\lambda$
$S$. $E \propto \frac{1}{d^3}$	$4$. Two infinite wires carrying uniform linear charge density parallel to the $x$-axis. The one along $(y=0, z=l)$ has a charge density $+\lambda$ and the one along $(y=0, z=-l)$ has a charge density $-\lambda$. Take $2l \ll d$
	$5$. Infinite plane with uniform surface charge density

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