Coulomb's law point charges distance for 0 net force

[jmr423]

Homework Statement

One charge of (+5µC) is placed in the air at exactly x = 0, and a second charge (+7µC) at x = 100cm. where can the third charge be placed so as to experience zero net force due to the other charges?

Homework Equations

F=KQ1Q2 / D^2

The Attempt at a Solution

Q1 = +5µC,
Q2 = +7µC,
Q3d from q1 = D1,
K = 9*10^9,
100 cm = 1 m
D1+D2=1m,
1m - D1 = D2.

K(q1*q3)/d1^2)=K(q2*q3)/(1-D1)^2
Then the k's cancel each other out and the Q3's cancel each other out.

(q1)/(d1^2)=(q2)/(1-D1)^2

Is this correct? also can you give me an example on solving this equation if it is correct?

Another thing
if Q1 had a smaller and opposite charge compared to Q2, the equation would be?
(q1)/(d1^2)=(q2)/(1+D1)^2

 

[jmr423]

Sorry i did not type out the values,

5/d^2=7/(1+D)^2

saying that we are finding distance and both the charges are 10^-6 it can cancel out right?
Sorry i guess this is more of a math problem if i have the formula right, i really am not sure how to solve this, i have tried but i keep getting the wrong answers. the answer in the book is .46m or 46cm

 

[ehild]

K(q1*q3)/d1^2)=K(q2*q3)/(1-D1)^2 Then the k's cancel each other out and the Q3's cancel each other out.

(q1)/(d1^2)=(q2)/(1-D1)^2

You meant D1 instead of d1, did you not? (q1)/(D1^2)=(q2)/(1-D1)^2 is correct. Cancel the factors 10^-6 . Take the reciprocal of both sides, expand the square, move everything to one side and solve the quadratic equation. You get two roots, exclude the one not between 0 and 1.

Another thing
if Q1 had a smaller and opposite charge compared to Q2, the equation would be?
(q1)/(d1^2)=(q2)/(1+D1)^2

Q1/D1^2=(q2)/(1+D1)^2 is the correct one.

 

[jmr423]

Thank you very much :P. I'm not 100% on solving quadratics however I have a 600 page math textbook that I'm working through right now that will cover it. My physics class is 1 year ahead of my math class. :S

 

[ehild]

Oh. Do not read all that 600 pages. If you have an equation of form ax²+bx+c=0 the solutions are

$$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ Can you manage?

ehild

 

[jmr423]

Oh. Do not read all that 600 pages. If you have an equation of form ax²+bx+c=0 the solutions are

$$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$ Can you manage?

ehild

Thank you very much, i will play around with it however i am not to familiar with radicals so I'm not sure if i will be able to get through it :S

 

[ehild]

Oh. Your Physics class is very much ahead of your Maths class. The square root of a number x is the non-negative number denoted by √x which multiplied by itself gives out x.

√x * √x = x.

√4=2 , as 2*2=4
√100=10 as 10*10 = 100.

You certainly find a key on your calculator that corresponds to the square root operator.



ehild