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

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The discussion focuses on solving an electric field problem involving two charges, 1.5x10^-6 C and 3.0x10^-6 C, separated by 0.20 m. The key steps involve setting the electric fields equal to each other and using the relationship between the distances from each charge to the point where the electric field is zero. Participants clarify that substituting the distance equation into the electric field equation is essential, and simplifying the algebra can lead to the correct quadratic form. The final quadratic equation derived is 0 = -1.5x10^-6(r1)^2 - 6x10^-7(r1) + 6x10^-8, which is confirmed to be correct.

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

Homework Statement



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

Homework Equations


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​
then quadratic forumula...

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.

 
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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.
 
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



final quadratic equals

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


but it's not right?
 
Hi viet_jon,

viet_jon said:
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



final quadratic equals

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


but it's not right?

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).)
 
yea your right...the answer is right...

I was comparing my answer to the one in the books, and the book's answer was in scientific notation, where mine was in full numbers...I overlooked it.thnkx guys...
 

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