Help Calculate Net Electric Force on Particle 3

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SUMMARY

The discussion focuses on calculating the net electric force on particle 3 in a system of four charged particles arranged in a square. The charges are Q1 = -Q2 = 100nC and Q3 = -Q4 = 200nC, with a separation distance of 5.0 cm. The user initially attempted to apply Coulomb's law using the formula F1,3 = (8.854x10^-12) x (Q1 x Q3)/0.0707xR^2 but expressed confusion over the complexity of their approach. A more straightforward method involves summing the forces in the x and y directions using vector components and the principle of superposition.

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Homework Statement


Hi, I would appreciate it if someone could help me with this question, I am a n00b here
Please and thank you
Okay here's the question.

The particles have charges Q1 = -Q2 = 100nC and Q3 = -Q4 =200nC
and distance a = 5.0cm.

What are the x and y components of the net electostatic force on particle 3 ?

So basically a diagram is given with the question showing the 4 particles equally separated from each other at a distance of 5.0cm, making a square. And I have to find the force on particle 3.

Homework Equations



This is the equation I used.

F1,3 = (8.854x10^-12) x (Q1 x Q3)/0.0707xR^2 (0.0707 comes from Phyagrous therom)
H= (square root) 0.05^2 + 0.05^2
F1,4 = (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2

Fnet = F1,3 + F1,4

= F1,3 + (8.854x10^-12) x (Q1 x Q3)/R^2 Cos 45 (45 degrees middle of square)
+ (8.854x10^-12) x (Q1 x Q4)/0.0707xR^2 Sin 45

The Attempt at a Solution



I have a feeling this is seriously wrong and have made it more complicated than it is.

I placed the values into the equation shown and came out with

= -79.46 x10^3 + 79.46 x10^3

Heres a site I found afterwards...
http://physics.bu.edu/~duffy/PY106/Charge.html

I think it might go a little something like that instead^^

If you could help that would be great thanks.
 
Last edited:
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Hey,

This is a rather straight-forward problem essentially all you need to do is sum up the forces on particle 3 in the x and y directions.

To get you started here is the general form for the superposition of forces (summing of force vectors).

<br /> \sum{\vec{F}_{m}} = {\sum_{i=1}^{n}}{\vec{F}_{mn}}<br />

So,

<br /> \sum{\vec{F}_{3_{}}} = {\vec{F}_{31_{}}} + {\vec{F}_{32_{}}}<br />

Breaking into the x-components,

<br /> \sum{\vec{F}_{3_{x}}} = {\vec{F}_{31_{x}}} + {\vec{F}_{32_{x}}}<br />

<br /> \sum{\vec{F}_{3_{x}}} = {\left(\frac{{k_{e}}{|q_{3}|}{|q_{1}|}}{{{r}_{31}}^{2}}\right)}{cos{\theta_{31}}}{\hat{i}} + {\left(\frac{{k_{e}}{|q_{3}|}{|q_{2}|}}{{{r}_{32}}^{2}}\right)}{cos{\theta_{32}}}{\hat{i}}<br />

From here you can use the symmetry of the problem to find your distances and evaluate.

Thanks,

-PFStudent
 
Last edited:
Thanks, those are the forumulas I used.

Made a silly few mistakes though
 

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