# Electric field of a flat metal plate and a point particle

I'm currently studying electric potential, and I'm having trouble with one of the problems on my homework:

A) A point particle with charge $+q$ is on the x-axis at a distance $d$ from the origin, and another point particle with charge $-q$ is on the x-axis at a distance $-d$ from the origin. Assuming the potential is 0 at large distances from these point particles, show that the potential is also 0 everywhere on the $x=0$ plane.

B) If an infinite grounded flat metal plate is on the $x=0$ plane, the negative charge is removed, and the electric potential is 0 in the same cases as in part (A), the potential function $V$ and the electric field $\vec{E}$ for this part is the same as in part (A). Using this result, find $\vec{E}$ at every point in the $x=0$ plane.

For part A, I simply did $V = \frac{kq}{r} + \frac{k(-q)}{r} = 0$, since the distance from the two particles to any point on the $x=0$ plane is the same for both particles.

However, this presents a problem with part B. What I found was $E = \frac{kq}{r^2} + \frac{k(-q)}{r^2} = 0$ using the same logic that I did for part A, which disagrees with the book's answer of $\frac{2kdq}{(d^2+r^2)^{3/2}}$. That looks like the formula for the electrical field of a ring, but I don't see the particular relevance here. What am I doing wrong?

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You are forgetting that the electric field is a vector quantity! Draw a diagram, and you will see that although the components on the $x=0$ plane of the electric fields of the two charges cancel each other, the components in the $x$-direction do not, and in fact they add and have the precisely the same magnitude and direction (hence the factor of 2). Also, be careful because
1. The $r$ is their formula is the distance to the origin of the point being considered, not the distance to the charge
2. When you find the field component in the $x$-direction, you need to make sure to multiply its magnitude by the appropriate $\cos\theta$ factor (again draw a diagram to see what you mean)