# Quick Question .I have the answer with steps, but can't figure out how th

1. May 7, 2008

### viet_jon

[SOLVED] Quick Question.....I have the answer with steps, but can't figure out how th

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

Two charges of 1.5x10^6C and 3x10^-6C are .20m apart. Where is the electric field between them equal to zero?

2. Relevant equations

3. The attempt at a solution

Variables

r(total)= .2m

q1= 1.5x10^-6

q2=3x10^-6

(r2)^2= (.20 - r1)^2

E1 = E2

kq1/(r1)^2 = kq2/(r2)^2

1.5x10^-6C/(r2)^2 = 3.0x10^-6C/(r2)^2

since
(r2)^2 = 2(r1)^2
substitude for (r2)^2 and rearrange:

0= (r1)^2 + .4r1 - 4.0x10^-2

the part I highlighted where it turns from red to blue, is the step I don't understand. I tried to rearrange it, but it doesn't come out as a quadratic as the book answer does.

2. May 7, 2008

### Tedjn

I believe your first red-colored step should be

1.5x10^-6C/(r1)^2 = 3.0x10^-6C/(r2)^2

Substituting $$r_2^2 = 2r_1^2$$ into your red equation is useless, because you used that equation to find the relationship in the first place. In the end, everything will cancel out, and you will get two equal numbers on both sides, which tells you nothing.

You need to use the other information you are given, that the total distance between the charges is 0.2 m. You had the right idea when you wrote (r2)^2= (.20 - r1)^2.

3. May 7, 2008

### viet_jon

I tried to substitude (r2)^2= (.20 - r1)^2 into the denominator of the right side....and got this

(1.5x10^-6) / (r1)^2 = (3.0x10^-6) / ( .2 - r1)^2

expand the blue i get

( (r1)^2 - .4r1 + 0.04)

then cross multiply I get

6x10^-8 - 6x10^-7r1 + 1.5x10^-6(r1)^2 = 3x10^-6(r1)^2

0 = -1.5x10^-6(r1)^2 - 6x10^-7(r1) + 6x10^-8

but it's not right?

4. May 8, 2008

### alphysicist

Hi viet_jon,

Why do you think it's not right? I think it gives the correct answer.

You can make the algebra a bit easier though. First you can cancel a factor of 10^-6 from each side of your equation

(1.5x10^-6) / (r1)^2 = (3.0x10^-6) / ( .2 - r1)^2

Also, you can take the square root of each side rather than expanding into a quadratic form.

(If you do this you'll want to make sure to write the factor in the denominator (such as the (.2-r1) that is being squared is positive. In other words, make sure the denominator is (.2-r1)^2 instead of (r1-.2)^2. If you put in the second form, the above equation is still correct, but when you take the square root you would need to put a minus sign in (because square root of a number can be negative).)

5. May 8, 2008