X and Y are large,parallel conducting plates close to each other. Each face has an area A. X is given a charge Q. Y is without any charge. Points A, B and C are as shown in the figure then:

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Answer 1

Charge on plate X is $Q_{1} = Q$ and charge on plate Y is $Q_{2} = 0$
Charges on plate surfaces 1 and 4 are $q_{1} = q_{4} = \frac{Q _{1} + Q _{2}}{2} = \frac{Q}{2}$
Charge on surface 2 is $q_{2} = \frac{Q _{1} - Q _{2}}{2} = \frac{Q}{2}$
and charge on surface 3 is $q_{3} = - q_{2} = - \frac{Q}{2}$
The field at A is $E_{A} = E_{1} + E_{4} = \frac{( Q / 2 )}{2 A ϵ _{0}} + \frac{( Q / 2 )}{2 A ϵ _{0}} = \frac{Q}{2 A ϵ _{0}}$
As both 1 and 4 surfaces have same positive charge so fields at A and C are directed away from the surfaces and have equal magnitude.so are equal in magnitude but opposite in direction.

Thus, $E_{A} = \frac{Q}{2 A ϵ _{0}}$ (toward left) and $E_{C} = \frac{Q}{2 A ϵ _{0}}$ (toward right)
Now $E_{B} = E_{2} + E_{3} = \frac{Q / 2}{2 A ϵ _{0}} + \frac{Q / 2}{2 A ϵ _{0}} = \frac{Q}{2 A ϵ _{0}}$ (toward right) as filled is directed from positive to negative charge.

Answer 2

Charge on plate X is

Q_{1} = Q

and charge on plate Y is

Q_{2} = 0

Charges on plate surfaces 1 and 4 are

q_{1} = q_{4} = \frac{Q _{1} + Q _{2}}{2} = \frac{Q}{2}

Charge on surface 2 is

q_{2} = \frac{Q _{1} - Q _{2}}{2} = \frac{Q}{2}

and charge on surface 3 is

q_{3} = - q_{2} = - \frac{Q}{2}

The field at A is

E_{A} = E_{1} + E_{4} = \frac{( Q / 2 )}{2 A ϵ _{0}} + \frac{( Q / 2 )}{2 A ϵ _{0}} = \frac{Q}{2 A ϵ _{0}}

As both 1 and 4 surfaces have same positive charge so fields at A and C are directed away from the surfaces and have equal magnitude.so are equal in magnitude but opposite in direction.

Thus,

E_{A} = \frac{Q}{2 A ϵ _{0}}

(toward left) and

E_{C} = \frac{Q}{2 A ϵ _{0}}

(toward right)

Now

E_{B} = E_{2} + E_{3} = \frac{Q / 2}{2 A ϵ _{0}} + \frac{Q / 2}{2 A ϵ _{0}} = \frac{Q}{2 A ϵ _{0}}

(toward right) as filled is directed from positive to negative charge.

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