# Finding how much to move a charged particle to obtain a certain net force on other charges

#### Fontseeker

I have the following problem:

So, I know the following:
$q_1 = -e$
$q_2 = -e$
$q_4 = -e$
$q_5 = e$
$e = 1.6*10^{-19}$
$k = 9*10^{9}$

Now, I can get the current $F_{net}$ of particle 5 with this information:

$F_{net} = (F_{3} - F_{1})\hat i + (F_{2} + F_{4})\hat j$
$F_{1} = k * \frac{e * e}{x_1^2} = 1.627 * 10^{-26}N$
$F_{2} = k * \frac{e * e}{y_2^2} = 6.508 * 10^{-26}N$
$F_{3} = k * \frac{e * e}{x_3^2} = 1.627 * 10^{-26}N$
$F_{4} = k * \frac{e * e}{y_4^2} = 1.627 * 10^{-26}N$

Now, I can break up $F_{net}$ into components:
$F_{x} = (1.627 * 10^{-26} - 1.627 * 10^{-26})N =0N$
$F_{y} = (6.508 * 10^{-26} - 1.627 * 10^{-26})N = 4.881 * 10^{-26}N$

To rotate $F_{net}$ by 30 degrees, I know the y component will not change, so I can make this assumption:
$\tan \theta= \frac{F_x}{F_{y}}$
so
${F_x} = \tan \theta * F_{y} = \tan 30 * 4.881 * 10^{-26}N = 2.818 * 10^{-26}N$
and since it is rotating counterclockwise, the x component must be negative:
${F_x} = -2.818 * 10^{-26}N$
Now, to find the new position, r, of particle 1 to get this new ${F_x}$ for the net charge of particle 5
${F_x} = {F_3} - {F_1}$
${F_1} = {F_3} - {F_x}$
${F_1} = {F_3} - {F_x}$
$k * \frac{e * e}{r^2} = {F_3} - {F_x}$
$r = \sqrt {\frac{k * e * e}{{F_3} - {F_x}}}$
$r = \sqrt {\frac{k * e * e}{1.627 * 10^{-26} - (-2.818 * 10^{-26})}}$
$r = \sqrt {\frac{k * e * e}{1.627 * 10^{-26} + 2.818 * 10^{-26}}}$
$r = 0.071995m$

So, the new position of particle 1 is at (0.071995m, 0), which is my answer for part A. Now, for part B, the new position of particle 3 would have to be (-0.071995m, 0). However, this is the wrong answer. Does anyone see my mistake?

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#### Delta2

Homework Helper
Gold Member
I believe by rotation it means rotation without changing the magnitude of the vector. In your calculations the magnitude of $F_{net}$ does not remain constant as it rotates. I believe the core of your mistake lies in this line
To rotate FnetFnetF_{net} by 30 degrees, I know the y component will not change, so I can make this assumption
I think the y component must also change so that the new $F_x$ and the new $F_y$ result in a vector that has the same magnitude as before.

#### Fontseeker

I believe by rotation it means rotation without changing the magnitude of the vector. In your calculations the magnitude of $F_{net}$ does not remain constant as it rotates. I believe the core of your mistake lies in this line

I think the y component must also change so that the new $F_x$ and the new $F_y$ result in a vector that has the same magnitude as before.
If the y component will also change, then I would have several unknowns. I wouldn't know the position of particle 1, particle 4, and particle 2.

#### Delta2

Homework Helper
Gold Member
Hm yes you are right, upon closer examination of the statement, it just states "to rotate the direction" and not to "rotate the vector"
Hold on while i check the rest of your work.

#### Delta2

Homework Helper
Gold Member
Hi, sorry for the late response, i really had to leave.

The only mistake I see is at the sign of the position of particle 1, it must be at $x=-0.071995$ while the new position of particle 3 at $x=0.071995$ if we take the x-axis to be positive at the right. You lost information about the sign because you took the positive square root, while there is the negative square root also that is $-\sqrt {K\frac{ee}{F_3-F_x}}$ which is also a valid solution to the equation. We have to choose the negative solution because only then the force from particle 1 to particle 5 is towards to the left and since we want the total force on x-axis to be towards to the left.

Another mistake i see that does not affect your end results is that since if we look at the statement for the $y_2$ and $y_4$ it should be
$F_2=1.627\times 10^{-26}$ and $F_4=6.508\times 10^{-26}$
Also since it seems that by $F_n$ $n=1,2,3,4$ you denote the magnitude of the force, the equation about $F_{net}$ should be
$\vec{F_{net}}=(F_3-F_1)\hat i+(F_4-F_2)\hat j$
However you denote as $F_x$ not only the magnitude but the direction since you give it a negative value , so you actually mixing things up denoting by $F_n$ the magnitude and by $F_x$ magnitude and direction. It should be better if you stated $\vec{F_x}=-2.818\times 10^{-26}\hat i$. But then again this doesn't affect your end result.

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#### Fontseeker

Hi, sorry for the late response, i really had to leave.

The only mistake I see is at the sign of the position of particle 1, it must be at $x=-0.071995$ while the new position of particle 3 at $x=0.071995$ if we take the x-axis to be positive at the right. You lost information about the sign because you took the positive square root, while there is the negative square root also that is $-\sqrt {K\frac{ee}{F_3-F_x}}$ which is also a valid solution to the equation. We have to choose the negative solution because only then the force from particle 1 to particle 5 is towards to the left and since we want the total force on x-axis to be towards to the left.

Another mistake i see that does not affect your end results is that since if we look at the statement for the $y_2$ and $y_4$ it should be
$F_2=1.627\times 10^{-26}$ and $F_4=6.508\times 10^{-26}$
Also since it seems that by $F_n$ $n=1,2,3,4$ you denote the magnitude of the force, the equation about $F_{net}$ should be
$\vec{F_{net}}=(F_3-F_1)\hat i+(F_4-F_2)\hat j$
However you denote as $F_x$ not only the magnitude but the direction since you give it a negative value , so you actually mixing things up denoting by $F_n$ the magnitude and by $F_x$ magnitude and direction. It should be better if you stated $\vec{F_x}=-2.818\times 10^{-26}\hat i$. But then again this doesn't affect your end result.
thanks

Homework Helper
Gold Member

#### Fontseeker

yes

"Finding how much to move a charged particle to obtain a certain net force on other charges"

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