Finding Electric Field of Two Charges

[erok81]

Homework Statement

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.

Homework Equations

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

Removing the test charge:

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

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?

 

[ehild]

Homework Statement

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.

Homework Equations

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

Removing the test charge:

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

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

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

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

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

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

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

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

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

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

You made a mistake. The red line is wrong.

ehild

 

[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?

 

[tiny-tim]

hi erok81!

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

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

hint: 1 = √q₁/√q₁ ​

 

[ehild]

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?

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

Spoiler

=5/3

is r=3/2 +1

Spoiler

=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