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Question

Three large identical conducting parallel plates carrying charge $+ Q, - Q$ and $+ 2 Q$ respectively, are placed as shown in the figure. If $E_{A}, E_{B}$ , and $E_{C}$ refer to the magnitudes of the electric fields at points $A, B$ and $C$ respectively, then

E_{A} > E_{B} > E_{C}

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E_{A} = 0

and

E_{C} > E_{B}

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E_{A} = 0

and

E_{B} > E_{C}

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E_{A} = 0

and

E_{B} = E_{C}

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Solution

The correct option is B $E_{A} = 0$ and $E_{C} > E_{B}$
$E_{1}, E_{2}$ and $E_{3}$ are electric fields due to plates $1, 2$ and $3$ respectively. Let $A$ is the surface area throught which the electric field lines pass perpendicularly.

Net electric field at point point $A$ :

$E_{A} = E_{1} + E_{2} - E_{3}$

Substituting the values of electric field from figure,

$E_{A} = \frac{Q}{2 ϵ_{0} A} + \frac{Q}{2 ϵ_{0} A} - \frac{2 Q}{2 ϵ_{0} A}$

$∴ E_{A} = 0$

Now net electric field at point $B:$

$E_{B} = E_{1} - E_{2} - E_{3}$

Substituting the values of electric field from figure,

$E_{B} = \frac{Q}{2 ϵ_{0} A} - \frac{Q}{2 ϵ_{0} A} - \frac{2 Q}{2 ϵ_{0} A}$

$∴ E_{B} = - \frac{Q}{2 ϵ_{0} A}$

Similarly, net electric field at point $C:$

$E_{C} = E_{1} - E_{2} + E_{3}$

Substituting the values of electric field from figure,

$E_{C} = \frac{Q}{2 ϵ_{0} A} - \frac{Q}{2 ϵ_{0} A} + \frac{2 Q}{2 ϵ_{0} A}$

$∴ E_{C} = \frac{Q}{ϵ_{0} A}$

Hence, $E_{C} > E_{B}$ and option (b) is correct answer.

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