Note that we didn’t plug in any negative signs! Rather, we calculated the magnitude of the electric field
produced by each charge, and showed the direction on the diagram.
Now, to find the net electric field at point P , we must add the electric field vectors. This is made
considerably simpler by the recognition that the y -components of the electric fields cancel ... both of
these vectors are pointed at the same angle, and both have the same magnitude. So, let’s find just the x -
component of one of the electric field vectors:
E (^) x = E cos θ , where θ is measured from the horizontal.
Some quick trigonometry will find cos θ ... since cos θ is defined as , inspection of the
diagram shows that . So, the horizontal electric field Ex = (510 m) ... this gives 140
N/C.
And now finally, there are TWO of these horizontal electric fields adding together to the left—one due
to charge “A” and one due to charge “B”. The total electric field at point P , then, is
280 N/C, to the left.
Part 2—Force
The work that we put into Part 1 makes this part easy. Once we have an electric field, it doesn’t matter
what caused the E field—just use the basic equation F = qE to solve for the force on the electron, where q
is the charge of the electron. So,
F = (1.6 × 10−19 C) 280 N/C = 4.5 × 10−17 N.