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erok81
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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
[itex]E=\frac{q_1 q_2 k_c}{r^2}[/itex]
Removing the test charge:
[itex]\frac{E}{q}=\frac{q_n k_c}{r^2}[/itex]
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
[itex]\frac{q_1 k_c}{r^2}=\frac{q_2 k_c}{(r-4^2)}[/itex]
[itex]\frac{(r-4)^2}{r^2}=\frac{q_2 k_c}{q_1 k_c}[/itex]
[itex]\left(\frac{(r-4)}{r}\right)^2=\frac{q_2 k_c}{q_1 k_c}[/itex]
[itex]\frac{r-4}{r}= \pm \sqrt{\frac{q_2}{q_1}}[/itex]
[itex]1-\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}}[/itex]
[itex]-1\left(-\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} -1\right)[/itex]
[itex]\frac{4}{r}= \pm \sqrt{\frac{q_2}{q_1}} +1[/itex]
[itex]\frac{r}{4}= \pm \sqrt{\frac{q_1}{q_2}} +1[/itex]
[itex]r= \left(\pm \sqrt{\frac{q_1}{q_2}} +1\right)4[/itex]
[itex]r= \pm 4\sqrt{\frac{q_1}{q_2}} +4[/itex]
[itex]r= 4 \pm 4\sqrt{\frac{1}{0.5}}[/itex]
[itex]r= 4 + 4 \sqrt{2}[/itex]
But that isn't correct. That isn't the correct answer. The correct answer is:
[itex]r= 8 + 4 \sqrt{2}[/itex]
Where does the extra factor of 2 come from making the 4 an 8?