Six infinitely large and thin non-conducting | vedclass

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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.

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