# Field in the case of two charged plates [duplicate]

A questions goes like this:

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.

To find the field at A, the solution goes; $$E_A = E_1 + E_4 = \frac{Q}{4A\epsilon_0}$$X$$2 = \frac{Q}{2A\epsilon_0}$$. The same is held true for the field at $$E_C$$.

How's this possible? If we label the surfaces from the left to the right as 1, 2, 3, and 4, wouldn't it be surface 3 that affects the field at A? And similarly, wouldn't surface 2 affect the field at C?

I think your question assumes that the plates are infinitely thin and are infinitely large, with plate $$X$$ having a certain surface charge density, $$\sigma$$. In other words, they don't have any thickness. Plate $$X$$ induces a negative charge density $$-\sigma$$ on plate $$Y$$ because of their close proximity.
If you assume the positive $$x$$-axis to be from left to right, the total field at point $$A$$ due to plates $$X$$ and $$Y$$ respectively is $$E_A = -\frac{\sigma}{\epsilon_0} + \frac{\sigma}{\epsilon_0} = 0.$$ By symmetry, the field at $$C$$ is zero as well (you can work this out if you want). At $$B$$, the fields add up constructively, resulting in $$E_B = +\frac{\sigma}{\epsilon_0} + \frac{\sigma}{\epsilon_0} = +2\frac{\sigma}{\epsilon_0}$$ toward the positive $$x$$-axis.