# Electric field between charges

1. Oct 19, 2007

### Ry122

Problem: There are two charges separated by a distance of one metre. Where between the two charges will the field strength be zero?
+5.4C ----------------- +4C

Attempt: Because they are both positive, the electric field lines diverge away from each other.
Does this mean that the electric field will be zero at 50cm?

2. Oct 19, 2007

### learningphysics

No, it's not 50cm. take an arbitrary point between the two charges... suppose it is a distance x from the left charge.

What is the total field at x?

set this field = 0... solve for x.

3. Oct 19, 2007

### Ry122

What formula do i use?

4. Oct 19, 2007

### learningphysics

E = kq/r^2

5. Oct 19, 2007

### Ry122

Since they are both positive and therefore repell each other, would it be correct to say that when their electrical fields are the same they would both negate each other and the resultant field would be zero?
Would this then be correct? kq/r^2=kq/r^2 Solve for r

6. Oct 19, 2007

### robphy

For the resultant field to be zero at that point, their field vectors must point in opposite directions, as well as have equal magnitude. On the line between the charges, the field vectors point in opposite directions... so now (as you suggest) you want to find when their magnitudes are equal.

What you want is
$$\]\displaystyle \frac{kq_1}{r_{\mbox{\small from q1}}^2} = \frac{kq_2}{r_{\mbox{\small from q2}}^2} \[$$
where $$r_{\mbox{\small from q1}}$$ and $$r_{\mbox{\small from q2}}$$ are the distances from the respective charges, which are somehow related to the separation between q1 and q2.

7. Oct 19, 2007

### Ry122

The equation simplified is 50400/x^2=37800/x^2
Is this solvable?

8. Oct 19, 2007

### learningphysics

if x is the distance of the point of interest from the left charge... then what is the distance of this point from the right charge?

the distance between the 2 charges is 0.50m. you're at a point x m to the right of the left charge (in between the 2 charges)... find the distance from the right charge... and that's what you use for r for the second charge...

not sure how you get 50400 and 37800... can you show your steps?

9. Oct 19, 2007

### Ry122

9*10^9 * (5.6*10^-6/x^2) = 9 * 10^9 * (4.2 * 10^-6/x^2)

10. Oct 19, 2007

### learningphysics

oh... ok. in your first post you wrote 5.4C and 4C.

fix the distances (see my last post)... then solve for x.

11. Oct 19, 2007

### Ry122

so u mean i should change 37800/x^2 to 37800/1-x^2
Since 20cm from the left charge would be 1-.2 from the right charge.
I still can't solve the equation.

12. Oct 19, 2007

### learningphysics

9*10^9 * (5.6*10^-6/x^2) = 9 * 10^9 * (4.2 * 10^-6/(1-x)^2)

50400/x^2 = 37800/(1-x)^2

solve for x. you have a quadratic in x.