Homework Help: Finding Electric Field of Two Charges

1. Feb 21, 2013

erok81

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

There is a charge Q=1 μC located at x=0 and a charge -Q/2 located at x=4 m. What is the E-field.

At what value of x is the E-field zero? Express your answer in meters and do not approximate decimals digits.

2. Relevant equations

$E=\frac{q_1 q_2 k_c}{r^2}$

Removing the test charge:

$\frac{E}{q}=\frac{q_n k_c}{r^2}$

3. The attempt at a solution

I have this solved, but I am going wrong somewhere and can't see my mistake. Here is my work and the correct final answer.

To find the E field at x=100, E1-E2=0. E2 is being subtracted due to q2 being negative.

This gives E1=E2

$\frac{q_1 k_c}{r^2}=\frac{q_2 k_c}{(r-4^2)}$

$\frac{(r-4)^2}{r^2}=\frac{q_2 k_c}{q_1 k_c}$

$\left(\frac{(r-4)}{r}\right)^2=\frac{q_2 k_c}{q_1 k_c}$

$\frac{r-4}{r}= \pm \sqrt{\frac{q_2}{q_1}}$

$1-\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}}$

$-1\left(-\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} -1\right)$

$\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} +1$

$\frac{r}{4}= \pm \sqrt{\frac{q_1}{q_2}} +1$

$r= \left(\pm \sqrt{\frac{q_1}{q_2}} +1\right)4$

$r= \pm 4\sqrt{\frac{q_1}{q_2}} +4$

$r= 4 \pm 4\sqrt{\frac{1}{0.5}}$

$r= 4 + 4 \sqrt{2}$

But that isn't correct. That isn't the correct answer. The correct answer is:

$r= 8 + 4 \sqrt{2}$

Where does the extra factor of 2 come from making the 4 an 8?

2. Feb 21, 2013

ehild

You made a mistake. The red line is wrong.

ehild

3. Feb 21, 2013

erok81

I am assuming I can't take the inverse of that thing like I did?

Would the more appropriate solution to cross multiply everything to get r alone?

4. Feb 21, 2013

tiny-tim

hi erok81!
that'll still leave you with the problem of what the inverse of the RHS is

hint: 1 = √q1/√q1

Last edited: Feb 22, 2013
5. Feb 21, 2013

ehild

You can not. Try it with numbers: if 1/r = 2/3 +1
=5/3
is r=3/2 +1
=5/2?
???

You can take the reciprocal of both sides of an equation if neither sides can be zero. If

4/r =a+b

then

r/4=1/(a+b)

Multiply both sides by 4:

r=4/(a+b)

Your equation is

$$\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} +1$$

Substitute the numerical values of the charges q2=Q/2 and q1=Q.

$$\frac{4}{r}= \pm \sqrt{1/2} +1$$

Take the reciprocal of both sides.

ehild

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