# Electric field of a plane

1. Jan 28, 2007

### threewingedfury

Two particles are fixed to an x-y coordinate system:
particle 1 of charge -5.00 microCoulombs lies on the x axis at x = + 6.00 cm and particle 2 of charge +5.00 microCoulombs lies on the y axis at y = + 8.00 cm. Midway between the particles, what is their net electric field, in unit vector notation?

So I got:
1/4piEo x q/r^2 - 1/4piEo x q/r^2

(1/(4pi(8.85x10^-12)) x (-5 x 10^-6)/(.06)^2)i - (1/(4pi(8.85x10^-12)) x (5 x 10^-6)/(.08)^2)j

-1.248e7i + 7024847j

Is this anywhere near right?

2. Jan 28, 2007

### Astronuc

Staff Emeritus
Not quite.

The electric field vector must point along the line from the positive charge to the negative charge, so one needs to determine the unit vector along that line.

The problem asks for the field at the midpoint between the charges, and so one needs the distance between the positive charge and the midpoint, and the negative charge and the midpoint. For one point at (x, 0) and the other at (0, y), think of (x/2, y/2).

3. Jan 28, 2007

### threewingedfury

so can anyone give me a better idea?

Last edited: Jan 29, 2007
4. Jan 29, 2007