Calculating Force of 5 MicroC Point Charge at (0, 0, 5) | Step-by-Step Guide

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To calculate the force on a 5 microC point charge located at (0, 0, 5) due to five identical 15 microC point charges positioned at the center and corners of a square defined by -1 < x < 1 and -1 < y < 1 on the z=0 plane, one must first determine the coordinates of the square's corners and center. The corners are located at (-1, -1, 0), (-1, 1, 0), (1, -1, 0), and (1, 1, 0), while the center is at (0, 0, 0). The force can be calculated using Coulomb's law, applying vector summation for the forces exerted by each charge. It is emphasized that integration methods are unnecessary for this problem due to the small number of point charges. Understanding the geometric distances and correctly applying the formulas will lead to the solution.
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Find the force ...Urgent!

1. Homework Statement [/b]

5 identical 15 microC point charges are located at the center and corners of a square which defined by -1 < x < 1, -1 < y < 1, z = 0.

Find the force on the 5 microC point charge at (0, 0, 5).



Stuck and Cannot Solve. Please Help !

How to calculate the distance ? i used the formula F=[(Q1)(Q2)(r-r1)(k)]/[(\sqrt{r-r1})^3]+...but i don't know what is the r for the coordinate since it = -1 < x < 1, -1 < y < 1, z = 0.

r=(x,y,z)...Sorry for my English
 
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Calculate the distances (geometrically) from the 6th particle to the others, then calculate numerically each force and finally add them all.
In general, try first with the distances, then the potential at (x,y,z), and finally the force.
 


Obama_US said:
Stuck and Cannot Solve. Please Help !

Tried Integrations Method...Any tips?

Please Help...I really cannot do !
Please show how far you got and which point you are stuck at.

Hint: You do not use any kind of integration method for a small number of point charges. Replace your integral with a vector sum.
 


Gokul43201 said:
Please show how far you got and which point you are stuck at.

Hint: You do not use any kind of integration method for a small number of point charges. Replace your integral with a vector sum.
this part "defined by -1 < x < 1, -1 < y < 1, z = 0." i cannot understand...the data is in variable...so how should i know what is the distance? cause it possible to be any value such as 0,0.1 and etc

Thanks for helping ...I just wanted to know use what method to solve
 
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Read the question carefully.

1. There are 5 point charges.
2. They are located in the (4) corners and (1) center of a square.
3. The square is defined by the region: x in the interval (-1,1), y in the interval (-1,1), z=0

First draw the x-y axes and draw the square that is described above. What are the co-ordinates of the corners and center of this square?
 
sorry i really don't get it...the co-ordinates of the center should be (0,0,0) but the corners co-ordinates i not sure cause it got range from -1 to 1...

correct me if i wrong

Thanks for helping again
 
The upper left corner is at (-1,1,0).
The upper right corner is at (1,1,0).
The first two numbers are like coordinates on graph paper.
 
i already solved this problem...thanks everyone who helping me
 
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