Two infinitely long parallel conducting plate | vedclass

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Question

(C) The electric field due to an infinitely large thin sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilo

Vedclass · Accepted Answer

$\frac{\sigma}{\varepsilon_0} 	ext{ V/m}$

Vedclass · Answer

(C) The electric field due to an infinitely large thin sheet of charge with surface charge density $\sigma$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.For two parallel plates with charge densities $+\sigma$ and $-\sigma$,the electric field from the positively charged plate points away from it,and the electric field from the negatively charged plate points towards it.In the region between the plates,both electric fields point in the same direction (from the positive plate to the negative plate).Therefore,the net electric field $E_{net}$ is the sum of the magnitudes of the individual fields:$E_{net} = E_1 + E_2 = \frac{\sigma}{2\varepsilon_0} + \frac{|-\sigma|}{2\varepsilon_0} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0} 	ext{ V/m}$.

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