# Electric Fields, Finding the zero field point

• Oblivion77
In summary, the problem asks for the distance along the line connecting two particles with positive charges q1 and q2 where the total electric field is zero. Using the equation E= \frac{kq}{r^2}, the distance x1 from q1 to the point where the field is zero is found to be a solution of the quadratic equation (q2-q1)x^2 + 2q1sx - q1s^2 = 0. By substituting back into the original equation, the solution can be checked. However, the solution does not cover all possible cases, such as when q1 = q2.
Oblivion77

## Homework Statement

Two particles with positive charges q1 and q2 are separated by a distance s. Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero.

## Homework Equations

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

## The Attempt at a Solution

Assuming x1 is the distance from q1 to the point where the field is 0.

This is what I get, I believe it is correct but I can't seem to solve for x1.

$$\frac{kq1}{x1^2}$$ = $$\frac{kq2}{(s-x1)^2}$$

Oblivion77 said:
This is what I get, I believe it is correct but I can't seem to solve for x1.

$$\frac{kq1}{x1^2}$$ = $$\frac{kq2}{(s-x1)^2}$$

Just "cross multiply" and expand:

$$\frac{kq_1}{x_1^2} =\frac{kq_2}{(s-x_1)^2}\implies kq_1(s-x_1)^2=kq_2 x_1^2$$

$$\implies q_1(s-x_1)^2=q_2 x_1^2 \implies q_1 x_1^2-2q_1 s x_1 +q_1 s^2=q_2 x_1^2$$

Then collect all terms on one side of the equations and solve the resulting quadratic.

Thanks, but I already got to that part above. How can I start to solve for x1 from that?
can I do this? (q2 - q1)x^2 + 2q1sx - q1s^2 = 0. What would I do from here?

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Oblivion77 said:
Thanks, but I already got to that part above. How can I start to solve for x1 from that?
can I do this? (q2 - q1)x^2 + 2q1sx - q1s^2 = 0. What would I do from here?

Yes you can do that.

Have you not learned the quadratic equation yet?!

Yea this is what it came out to, which is wrong unfortunately. Is there something I can simplify? I also tried the "+" instead of the "-", no luck.

Oblivion77 said:
Yea this is what it came out to, which is wrong unfortunately.
Check your reference: you got the quadratic formula wrong. (Were you using a computer algebra system? It looks like you forgot a pair of parentheses)

Incidentally, in order to try simplifying things, did you think about expanding things first?

Hurkyl said:
Check your reference: you got the quadratic formula wrong. (Were you using a computer algebra system? It looks like you forgot a pair of parentheses)

Incidentally, in order to try simplifying things, did you think about expanding things first?

I was doing it by hand =), I will try and expand and see if anything simplifies. Thanks.

Easier method will be to take the square root of the equation you have written and solve for x1. If the charge are same take + sign. If they are opposite take - sign.

Ok, after doing it again I got this? Does it look better? Stuff indeed canceled under the square root sign.

That looks plausible. (Note things still simplify more!) You can always try substituting back into the original equation... (which is usually a very good idea: there is rarely a better way to check if you've found a solution to an equation than actually plugging it in)

Incidentally, how did you decide that was the solution that lies on "the line connecting the two charges" (which I assume meant to say 'line segment'), rather than the other solution?

It turns out this answer is acceptable but it does simplify more. I chose the + instead of the - in the formula because the distance is somewhere between the 2 charges. I assumed using the "-" would be somewhere left of q1

Oblivion77 said:
I assumed using the "-" would be somewhere left of q1
Don't assume, prove! The argument isn't difficult, once you know what the right idea is. But you have to make note of your assumptions, and cover all of the cases... what if q1 > q2?

By the way, I note that your derivation (and your solution) isn't complete; it doesn't cover all possible cases... in particular, what if q1 = q2?

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## 1. What is an electric field?

An electric field is a physical phenomenon that is created by electrically charged particles and can be felt as a force on other charged particles within the field.

## 2. How do you calculate the strength of an electric field?

The strength of an electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge itself. This is known as Coulomb's law.

## 3. What is the direction of an electric field?

The direction of an electric field is the direction in which a positive test charge would move if placed in the field. Electric field lines are used to represent the direction of the field, with the lines pointing away from positive charges and towards negative charges.

## 4. What is the purpose of finding the zero field point?

The zero field point is the point in space where the electric field strength is zero. It is important to find this point to understand the behavior and characteristics of the electric field, such as its direction and shape.

## 5. How do you find the zero field point?

The zero field point can be found by setting the electric field equation equal to zero and solving for the coordinates of the point. This can be done mathematically or visually by plotting electric field lines and identifying where they intersect or form a null point.

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