# Coulomb Forces

## Homework Statement

Find the force on a point of charge of 30μC at (0,0,5)m due to a 4m square int he z=0 plane between x=±2m and y=±2m with a total of 500μC, distributed uniformly.

## The Attempt at a Solution

R=-x,-y,+5z
dQ=ρdydx=500/4 dydx

dE=dQ(5az) / (4πε0((x2+y2+25)3/2 )
x and y axis cancel

How to do integral with x2 and y2 in the denominator?

Ans. 4.66azN

Last edited:

Simon Bridge
Homework Helper
... use the relationship of dQ to dx and dy to make a double integral.

• azizlwl
... use the relationship of dQ to dx and dy to make a double integral.
... use the relationship of dQ to dx and dy to make a double integral.

Actually I really don't understand the question. Area enclosed by x=±2 and y=±2 will be 16m square. What does it means by" due to a 4m square "?

gneill
Mentor
Actually I really don't understand the question. Area enclosed by x=±2 and y=±2 will be 16m square. What does it means by" due to a 4m square "?
Draw a square 4 m per side. Distribute 500 μC uniformly on the edges (so each side of the square is a line segment with 500/4 μC on it.

F1d=3x5x9/16 ∫-22 dy (√29d -yay)/(29+y2)3/2
F1d=8.43 x 0.129 =1.087
The integral value using app integral calculator.
F1z=1.087 x 5 /√29 = 1.009.

Total force= 4Naz less than the answer..

gneill
Mentor
F1d=3x5x9/16 ∫-22 dy (√29d -yay)/(29+y2)3/2
F1d=8.43 x 0.129 =1.087
The integral value using app integral calculator.
F1z=1.087 x 5 /√29 = 1.009.

Total force= 4Naz less than the answer..
I don't understand what your last line is trying to convey. Can you explain in more detail?

4 segments, so I added all that z-axis component. X and y axis cancelled. The answer is 4.66azN

gneill
Mentor
So your result is about 4 N and the answer given in the text is 4.66 N ?

I'm inclined to agree with your result over that of the book.

• azizlwl