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Question

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

Vedclass · Accepted Answer

$+\frac{2q\sigma}{\varepsilon_0} \hat{k}$

Vedclass · Answer

(A) The electric field due to an infinitely long charged sheet with surface charge density $\sigma$ is given by $\vec{E} = \frac{\sigma}{2\varepsilon_0} \hat{n}$,where $\hat{n}$ is the unit vector normal to the sheet pointing away from it.At point $P$ (located between $z=a$ and $z=2a$):$1$. Due to the sheet at $z=2a$ with charge density $+\sigma$: $\vec{E}_1 = -\frac{\sigma}{2\varepsilon_0} \hat{k}$ (directed downwards).$2$. Due to the sheet at $z=a$ with charge density $-2\sigma$: $\vec{E}_2 = -\frac{2\sigma}{2\varepsilon_0} \hat{k} = -\frac{\sigma}{\varepsilon_0} \hat{k}$ (directed towards the sheet,i.e.,downwards).$3$. Due to the sheet at $z=-a$ with charge density $-\sigma$: $\vec{E}_3 = -\frac{\sigma}{2\varepsilon_0} \hat{k}$ (directed towards the sheet,i.e.,downwards).The net electric field at $P$ is $\vec{E}_{net} = \vec{E}_1 + \vec{E}_2 + \vec{E}_3 = -\left(\frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{\varepsilon_0} + \frac{\sigma}{2\varepsilon_0}\right) \hat{k} = -\frac{2\sigma}{\varepsilon_0} \hat{k}$.The electric force on charge $-q$ is $\vec{F} = (-q) \vec{E}_{net} = (-q) \left(-\frac{2\sigma}{\varepsilon_0} \hat{k}\right) = +\frac{2q\sigma}{\varepsilon_0} \hat{k}$.

Three infinitely long charged sheets are placed as shown in the figure. The electric force acting on a charge $-q$ placed at the point $P$ is ($\sigma=$ surface charge density,$\varepsilon_0=$ permittivity of free space).

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