# Electric Field due to Two Point Charges

1. Jan 23, 2008

### cse63146

1. The problem statement, all variables and given/known data

Two point charges are placed on the x axis. The first charge, q1= 8 nC, is placed a distance 16m from the origin along the positive x axis; the second charge, q2= 6 nC , is placed a distance 9m from the origin along the negative x axis.

Find the electric field at the origin, point O.

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

Now, assume that charge q2 is negative; q2 = -6 nC. What is the net electric field at the origin, point O?

Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. Keep in mind that an x component that points to the right is positive and a y component that points upward is positive.

2. Relevant equations

E = $$\frac{kQ}{r^2}$$

3. The attempt at a solution

I start by "placing" a test charge at the origin.

Here's what I think I should do:

E = $$\frac{kQ_1}{r^2}$$ + $$\frac{kQ_2}{r^2}$$

E_1 = $$\frac{(9x10^9)(8x10^-9)}{16^2}$$ = 0.261 N/C

E_2 = $$\frac{(9x10^9)(6x10^-9)}{9^2}$$ = 0.667 N/C

E = E_1 + E_2 = 0.261 + 0.667 = 0.928 N/C

Am I right?

Last edited: Jan 23, 2008
2. Jan 23, 2008

### Midy1420

The direction of the E-field is taken to be the direction of the force it would exert on a positive test charge so when you draw a diagram with a positive test charge how would the forces from the q1 and q2 be acting on it? In the same direction or opposite?

3. Jan 23, 2008

### cse63146

Since all of the charges are positive, they would repel, and thus create a force pushing it to the left.

Still not sure (electircity isnt my strong point)

4. Jan 23, 2008

### Midy1420

q1 is to the right of the test charge (placed at the origin) and q2 is to the left of it.. draw a force diagram to see

Last edited: Jan 23, 2008
5. Jan 23, 2008

### cse63146

ohhhh.... I thought the second charge was also placed on the positive x-axis

so would it be 0.667 - 0.261 since the force is pushing the test charge to the left?

and I just noticed that I posted it in the wrong section. Would someone be able to move it to intro physics?

Last edited: Jan 23, 2008