9/9/09

I supplemented Problem 51 to practice some vector handling skills. In the original problem, there are three charges: -4nC at (-0.5, 0); 5 nC at (0, 0) and 3 nC at (0.8, 0). The problem asks for electric field at (2,0). The added parts are:

(b) Calculate the electric field at (0,-2).

(c) Where, on the x axis is the electric field zero.

Solution;

A. Calculate the magnitude without any account of the charge signs. Use the charge sign to determine the field direction. Add them to get the total field.

For part a, the two positive charges will produce a field pointed away from them in the +x direction. The negative charge will give rise to a field in the –x direction.

N/C

B.Now consider the point (0, -2). The vector directions of the electric field at the observation point are shown. They should be added as vectors to get the net field. On the side, you see a display of putting the vectors tails to head and the resulting (no color arrow) sum. In calculation it is better to add the components.

The x component of the total field Ex = E3x + E4x+E5x where E3x is the x component of the field due to the charge 3nC and E5x = 0.

Here the cosines can be calculated from the larger triangles and give the square roots in the denominator. If you have doubts about how to do that, ask me and I will show you.

The y component has contributions from all three charges.

The magnitude

Here the arc tan gives 63º but since both Ex and Ey are negative you have to add 180° .

  1. To find the location where the electric field is zero: This is rather hard to calculate. A simpler question can be to divide the x axis into four regions:
  1. x>0.8: The electric field in this region can not be zero (except at infinity) since the positive charges are closer, with an E along positive x direction. They overpower the –x direction contribution from the weaker and farther away negative charge.
  2. 0.8> x > 0: In this region, there is a solution. The contributions from 3nC and 5nC charges are in the opposite direction. The effect of -4nC charge would be to move the location of the E = 0 point to the left. The strongest contributions come from the two positive charges which cancel at x0 =0.45cm. The effect of -4nC charge is to move that point a little to the left.
  3. 0> x > -.5: In this region, the electric fields due to all charges are all in the –x direction. It is not possible to find a zero here.
  4. x < -0.5: Here cancellation is possible. -4nC is close and the electric field is pointed to right. The positive charges provide a field pointed to the left. The strongest contenders for contribution are -4nC and 5nC. Where is the electric field zero if they were the only two charges around?

The solutions exist in regions II and IV. It may be advisable to solve the equations numerically to get to the answer. If you do that you will see how close you could have guessed the answer.