Electric field due to a charged square

In summary: L in one of the expressions.In summary, a square has a charge of Q and has a field that is 0 when z is 0 and goes to 4 when z goes to infinity.
  • #1
WrongMan
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Homework Statement


there is a square on the XoY plane, centered at the origin (just outlines of the square) it has a charge Q (Q>0) and side 2L, i must evaluate the electric field along the z axis. see attached image

Homework Equations


E=k*q/r^2

The Attempt at a Solution


So first i divided the square into 4 lines, and noticed that due to the symmetry there is only a field in the z direction, and all the lines of the square contribute the same to that field.

So:
dE= k(Q/8L)1* dx/(x2+y2+z2)2*(z/(x2+y2+z2))½)3

where subscrip1 is Q/8L because the full Q is for the whole square, so we divide it by 4 to get the charge on the line and divide again by 2L to get the charge density.
subscript 2 is the distance of the charge to the line.
subscript 3 is cosine of E with Ez so we get only the z component
Now the integral, i get an not-so-straightforward integral, so if there is a simplification i could do here please advise.
so i get
E= k(Q/8L)z*∫dx/((x2+y2+z2)3/2))
and i integrate either over -L to L or from 0 to L and multiply it by 2.
and i get

E=kQz*1/((y2+z2)*(L2+y2+z2)½)

This result makes sense to me physically, as field is 0 when z is 0 (makes sense since field would no longer have a z component) and also goes to 0 when z goes to infinity.

So the total field is 4*E. Am i correct? any simplification suggestions for the integral

EDIT: oh and y should be L since its fixed.
 

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  • #2
WrongMan said:
E= k(Q/8L)z*∫dx/)(x2+y2+z2)*(x2+y2+z2)3/2))
Did you intend to leave out the part that is marked in blue?
and i integrate either over -L to L or from 0 to L and multiply it by 2.
and i get

E=kQz*1/(y2+z2)*(L2+y2+z2))½)
Note that your parentheses don't match in this expression.
Did you already multiply by 4 to take into account the 4 sides? If so, I believe you are getting the correct answer (with y = L).
any simplification suggestions for the integral
I don't see anything that would simplify what you have done.
 
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  • #3
TSny said:
Did you intend to leave out the part that is marked in blue?

Note that your parentheses don't match in this expression.
Did you already multiply by 4 to take into account the 4 sides? If so, I believe you are getting the correct answer (with y = L).

I don't see anything that would simplify what you have done.

Im sorry about the confusion with the expressions I've been trying to find a way to write them properly here (sugestions??).
About your first quote
yes i did, i confused myself when writing the expressions here.

about your second quote, everything after "1" is supposed to be in the denominator (bellow "1"), again, confusion when writing the expressions here.
and i multiplied by 4 bellow.

im fixing the expressions.
So it's correct?
 
  • #4
WrongMan said:
So it's correct?
Your method is correct.

Can you write out your final expression for the answer?
 
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  • #5
TSny said:
Your method is correct.

Can you write out your final expression for the answer?

E= 4*(kQz*1/((L2+z2)*(L2+L2+z2)½)) in the z direction

EDIT: i replaced y for L since when i evaluated the single line y didnt vary, it was fixed at y=L and i evaluated the change as x went from -L to L
 
  • #6
WrongMan said:
E= 4*(kQz*1/((L2+z2)*(L2+L2+z2)½)) in the z direction
I don't think the factor of 4 is correct.
 
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  • #7
TSny said:
I don't think the factor of 4 is correct. And of course you should replace y by its value in terms of L.
So i eavluated the field created by one line, assuming the opposite line would counter the field in the direction from one line to the oposite one, and won't all 4 lines contribute the same to the field?
 
  • #8
WrongMan said:
So i eavluated the field created by one line, assuming the opposite line would counter the field in the direction from one line to the oposite one, and won't all 4 lines contribute the same to the field?
Yes, all four sides will contribute equally. But, before multiplying by 4 to take care of the 4 sides, I get a 4 in the denominator for the expression due to 1 side alone. So, the 4's cancel in getting the final answer.
 
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  • #9
One way to check the overall numerical factor in your answer is to look at the limit where L → 0 (i.e., L << z). The square should "act like a point charge" in this limit.
 
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  • #10
TSny said:
Yes, all four sides will contribute equally. But, before multiplying by 4 to take care of the 4 sides, I get a 4 in the denominator for the expression due to 1 side alone. So, the 4's cancel in getting the final answer.
OHHH yeah you are right.
in solving the problem, i re-did it saying the line had charge Q (to simplify a bit).
As you might notice the Q/8L dissapears from one expression to the next, Q/8L should simplify with the 2L from the integral becoming Q/4 and then when multiplied by 4 they'd cancel out. my bad, i get distracted :/
 
  • #11
OK. Good work!
 
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Related to Electric field due to a charged square

What is an electric field?

An electric field is a physical field that describes the influence that electrically charged objects have on each other. It is a vector field, meaning it has both magnitude and direction, and is created by the presence of electric charges.

How is the electric field due to a charged square calculated?

The electric field due to a charged square can be calculated using Coulomb's law, which states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. The equation for electric field due to a charged square takes into account the size of the square, the magnitude and direction of each charge, and the distance from the square to the point where the electric field is being measured.

What factors affect the strength of the electric field due to a charged square?

The strength of the electric field due to a charged square is affected by several factors, including the magnitude and direction of the charges on the square, the distance from the square to the point where the electric field is being measured, and the size of the square itself. Additionally, the presence of other charged objects in the surrounding area can also impact the strength of the electric field.

What is the direction of the electric field due to a charged square?

The direction of the electric field due to a charged square depends on the relative positions and charges of the square and the point where the electric field is being measured. Generally, the electric field lines will point away from positively charged objects and towards negatively charged objects. In the case of a charged square, the direction of the electric field will vary depending on the position of the point in relation to the square and the distribution of charges on the square.

Can the electric field due to a charged square be negative?

Yes, the electric field due to a charged square can be negative. This occurs when the net charge of the square is negative, meaning there are more negative charges than positive charges, and the point where the electric field is being measured is closer to the square than any of the individual charges. In this case, the electric field will be directed towards the square and have a negative magnitude.

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