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Question

(D) The electric field due to an infinite thin sheet of charge with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\epsilon_0}

Vedclass · Accepted Answer

$\vec{E}_{ I }=-\frac{\sigma}{\epsilon_0} \hat{n}, \vec{E}_{ II }=0, \vec{E}_{ III }=\frac{\sigma}{\epsilon_0} \hat{n}$

Vedclass · Answer

(D) The electric field due to an infinite thin sheet of charge with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\epsilon_0} \hat{n}$,where $\hat{n}$ is the unit vector normal to the sheet pointing away from it.Let the direction to the right be the positive direction $\hat{n}$.In region $I$ (to the left of both sheets),both sheets produce an electric field pointing to the left $(-\hat{n})$:$\vec{E}_{ I } = -\frac{\sigma}{2\epsilon_0} \hat{n} - \frac{\sigma}{2\epsilon_0} \hat{n} = -\frac{\sigma}{\epsilon_0} \hat{n}$.In region $II$ (between the two sheets),the left sheet produces a field to the right $(+\hat{n})$ and the right sheet produces a field to the left $(-\hat{n})$:$\vec{E}_{ II } = \frac{\sigma}{2\epsilon_0} \hat{n} - \frac{\sigma}{2\epsilon_0} \hat{n} = 0$.In region $III$ (to the right of both sheets),both sheets produce an electric field pointing to the right $(+\hat{n})$:$\vec{E}_{ III } = \frac{\sigma}{2\epsilon_0} \hat{n} + \frac{\sigma}{2\epsilon_0} \hat{n} = \frac{\sigma}{\epsilon_0} \hat{n}$.

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions $I, II$ and $III$ are:

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