# Electric field outside of space between two plates of identical, yet opposite charge

Can somebody clarify the following statement:

Two plates, with the same but opposite charge, are placed opposite each other. The electric field between them is equal to σ/2εo - (-σ/2εo) = σ/εo However, the electric field outside of the space between the two plates, for example at point A, located halfway the positive plate, is equal to σ/2εo - σ/2εo = 0.

There is no drawing to illustrate the statement.

I know the formula σ/2εo + σ/2εo = σ/εo, and understand how this can be bent into σ/2εo - (-σ/2εo) = σ/εo, as they are opposite charges, and need to be added, as they reinforce each other.

However, I am lost as to σ/2εo - σ/2εo = 0 for the external point. Could somebody clarify the latter bit?

Thank you!

• – Frobenius Aug 27 '18 at 12:25

Referring to the included figure, you can see the fields due to the +ve charged plate: These fields (in red) are directed away from that plate and of strength $\sigma/2 \epsilon_0$. For the -ve charged plate the fields (in blue) are directed towards the plate and of strength $-\sigma/2 \epsilon_0$.
Between the plates the fields to the $\pm$ plates add to each other, while outside they cancel.