Each of two large conducting parallel plates | vedclass

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(B) When a charge $Q$ is given to one of two large parallel conducting plates,the charge redistributes itself on both sides of each plate such that th

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$\frac{Q}{2 A \varepsilon_0}$

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(B) When a charge $Q$ is given to one of two large parallel conducting plates,the charge redistributes itself on both sides of each plate such that the electric field inside the conducting material is zero.Let the charges on the four surfaces (from left to right) be $q_1, q_2, q_3,$ and $q_4$.For the first plate: $q_1 + q_2 = Q$ and for the second plate: $q_3 + q_4 = 0$.Since the plates are large,the electric field due to a surface with charge density $\sigma$ is $E = \frac{\sigma}{2 \varepsilon_0} = \frac{q}{2 A \varepsilon_0}$.Inside the conducting material of the plates,the net electric field must be zero.Applying this condition,we find that the charges on the inner surfaces are $q_2 = \frac{Q}{2}$ and $q_3 = -\frac{Q}{2}$,while the outer surfaces have $q_1 = \frac{Q}{2}$ and $q_4 = \frac{Q}{2}$.The electric field at a point between the plates is due to the inner surface charges $q_2$ and $q_3$.$E_{	ext{net}} = E_2 + E_3 = \frac{q_2}{2 A \varepsilon_0} + \frac{|q_3|}{2 A \varepsilon_0} = \frac{Q/2}{2 A \varepsilon_0} + \frac{Q/2}{2 A \varepsilon_0} = \frac{Q}{4 A \varepsilon_0} + \frac{Q}{4 A \varepsilon_0} = \frac{Q}{2 A \varepsilon_0}$.

Each of two large conducting parallel plates has one-sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral,then the electric field at a point in between the plates is given by

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