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Question

(B) The electric field due to an infinitely long charged sheet with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2 \epsilon_0}

Vedclass · Accepted Answer

$2$

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(B) The electric field due to an infinitely long charged sheet 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 sheets be at $x = -a$,$x = a$,and $x = 3a$ with surface charge densities $-\sigma$,$-2\sigma$,and $\sigma$ respectively.Point $P$ is located between $x = a$ and $x = 3a$.$1$. Electric field due to the sheet at $x = -a$ (charge density $-\sigma$): The field points towards the sheet (in the negative $x$-direction). $\vec{E}_1 = \frac{|-\sigma|}{2 \epsilon_0} (-\hat{i}) = -\frac{\sigma}{2 \epsilon_0} \hat{i}$.$2$. Electric field due to the sheet at $x = a$ (charge density $-2\sigma$): The field points towards the sheet (in the negative $x$-direction). $\vec{E}_2 = \frac{|-2\sigma|}{2 \epsilon_0} (-\hat{i}) = -\frac{2\sigma}{2 \epsilon_0} \hat{i} = -\frac{\sigma}{\epsilon_0} \hat{i}$.$3$. Electric field due to the sheet at $x = 3a$ (charge density $\sigma$): The field points away from the sheet (in the negative $x$-direction). $\vec{E}_3 = \frac{\sigma}{2 \epsilon_0} (-\hat{i}) = -\frac{\sigma}{2 \epsilon_0} \hat{i}$.The net electric field at point $P$ is $\vec{E}_P = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 = (-\frac{\sigma}{2 \epsilon_0} - \frac{\sigma}{\epsilon_0} - \frac{\sigma}{2 \epsilon_0}) \hat{i} = -\frac{2\sigma}{\epsilon_0} \hat{i}$.The magnitude of the electric field is $|\vec{E}_P| = \frac{2\sigma}{\epsilon_0}$.Comparing this with $\frac{x \sigma}{\epsilon_0}$,we get $x = 2$.

Three infinitely long charged thin sheets are placed as shown in the figure. The magnitude of the electric field at point $P$ is $\frac{x \sigma}{\epsilon_0}$. The value of $x$ is . . . . . . . (All quantities are measured in $SI$ units).

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