- #1

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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?

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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