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# Electric field?

I got .314 for part A but I'm being told it's wrong. I know that the answer for part A has to be closer to 0 than 3. I know I need to set the Electric field to 0, but after that, I'm at a lost for what to plug in.

Two charges, -20 and +4.2 µC, are fixed in place on the x axis at x = 3.0 m and x = 0 m, respectively.

(a) At what spot along the x axis is the net electric field zero?

x = m

(b) What would be the force on a charge of +19 µC placed at this spot?

N

You're English is great, and thanks for trying to help. I typed that answer into my homework website and it says it is wrong. :-(

### 2 Answers

- Dorian36Lv 41 decade agoFavorite Answer
First, apologizing for my not so perfect English...

We have to recall what is the meaning of electric field: it is a vector quantity, with a direction of the electric force on so called test charge, this is a positive point charge. Since the first charge (at x = 3.0 m) is positive and the other one (at the origin) is negative, they both act on the test charge, placed between them on the x-axis, toward right (the first charge reppels the test one, the second charge attracts). So, we found out that the net electric field is zero somewhere on the x-axis between two charges. Let's say it is in point A.

Than, we write equation: E1 = E2.

We write the distance between the first charge and the point A as x, giving the distance between A and the second charge r - x (where r = 3 m).

We remember the equation for the magnitude of electric field of a point charge: E = q / (4*pi*epsilon 0*r^2). We have q1 = +4.2*10^-6 C and q2 = -20*10^-6 C and we can write now:

q1 / (4*pi*epsilon o*x^2) = q2 / (4*pi*epsilon 0*(r - x)^2).

All the constants on both sides cancel and we have:

q1 / x^2 = q2 / (r - x)^2.

If we solve for x, it would give us quadratic equation for x, but we can avoid this by using square-root on equation:

sqrt(q1) / x = sqrt(q2) / (r - x)

or

sqrt(q1 / q2) = x / (r - x).

If we solve for x, we get x = sqrt(q1 / q2)*r / (1 + sqrt(q1 / q2)) = 0.943 m.

Yes, the answer is closer to 0 than 3 because the magnitude of the positive charge at the origin is almost 5-times smaller than the magnitude of the negative one.

I hope this answers your question :-)

- lavzonLv 44 years ago
electric field E= - dV /dr As skill is consistent dV / dr =0 for this reason electric field is 0 in a area the place electric skill is consistent. case in point, interior a hollow charged matallic sphere, the electrical powered field IS 0 however the flair interior is consistent and equivalent to skill on the suface of charged sphere