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

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  • #1
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I have the following problem:

Screen Shot 2019-08-30 at 6.41.21 PM.png

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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  • #2
Delta2
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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.
 
  • #3
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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.
 
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  • #4
Delta2
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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.
 
  • #5
Delta2
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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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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
 
  • #7
Delta2
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thanks
What does your text book answer key says, is x=-0.071995 the correct answer (for particle 1)?
 
  • #8
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What does your text book answer key says, is x=-0.071995 the correct answer (for particle 1)?
yes
 
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