# Homework Help: Where does electric field go to zero?

1. Jun 8, 2014

### baby_1

Hello
Here is an example.we want to calculate value of x where Electric filed going to be zero.i want to use vector form of coulomb equation

So as we know

for 1c:

for -9C:

so as we do superposition we obtain

So

In this problem x doesn't have any real number.what is my problem?

Any help appreciate

2. Jun 8, 2014

### Jilang

Why does your second equation have a -ax, but your third equation doesn't?

3. Jun 8, 2014

### CAF123

The electric field due to the two charges should be pointing in different directions at some position x to the left of the 1C charge. This condition is not met in the equation where you superimpose the fields.

4. Jun 8, 2014

### baby_1

Hello
Dear Jilang
Electric filed in equation 2 is in (-ax) direction but the third equation is in (+ax) direction it is because of charges positive or negative value.

dear CAF123
as i check the book answer that use only coulomb scale form,EF pointing and where electric filed going to be zero is in near 1c charge but outside of two charges distribution.

but i want to use vector form of coulomb equation.

Last edited: Jun 8, 2014
5. Jun 9, 2014

### CAF123

In your coordinate system, the E field from the 1C charge at a distance x defined in the sketch is $E_{1C} = - kq_1/x^2 \underline{e}_x$. That of the -9C charge is $E_{-9C} = kq_2/(10+x)^2 \underline{e}_x$. $q_1, q_2$ here are the magnitudes of the charges and their differing sign is manifest in the reversal of the unit vector.

6. Jun 9, 2014

### baby_1

Hello
Dear CAF123
however i don't agree with your explanation it is because of charges act.for example we know that the electric filed line of a dot positive charge is :

and EF equation is :

but if we put a dot negative charge that EF line is same as here

we can only change the sign of charge and have a correct form.

now why the vector form of coulomb equation doesn't work here?

7. Jun 10, 2014

### CAF123

That is the same as $$\vec E = \frac{Q}{4 \pi \epsilon_o r^2} (- \hat r) = -\frac{Q}{4 \pi \epsilon_o r^2} \hat r$$ If you look at the picture, the electric field due to a point charge is spherically symmetric and the radial vector is defined to be positive radially outwards (spherical coordinates convention). In the case of a negative charge, the field lines move in the radially inwards direction, which is exactly what $-\hat r$ in the above equation is telling you.

8. Jun 10, 2014

### Jilang

Hi Baby_1 , your post #6 is correct. But the r vectors both go the same way, pointing to the left of both charges in the negative x direction.

9. Jun 10, 2014

### baby_1

Thanks dear CAF123 & jilang for your response.
I found my problem.i wrote wrong the distance vector.here is an Electromagnetic book that explain better

here is the correct answer :

10. Jun 10, 2014

### BiGyElLoWhAt

Is that the correct answer or correct equation?

Try this:

You know the electric field is 0 at some point.

This means that $\sum E_{q} = ?$

Expand the sum and you have an equality to work with that's fairly easy to solve, and you're "using" the vector form of columbs law. Really, in this situation, the vectors are only useful for telling you the direction of the contribution to the field at a point by the respective point charge.

11. Jun 10, 2014

### BiGyElLoWhAt

also I think you meant $K\frac{9}{(x+10)^2}a\hat{x}$

12. Jun 10, 2014

### Jilang

Well done Baby_1!