Electric Field on finite charged rod

In summary, a thin rod with length L=100 cm and total charge q0=37 nC has a line density \lambda(x) =kx^2 that is proportional to the square of the distance from the end A. To determine the electric field E at the end A, we must integrate over the length of the charged rod and solve for k. The formula E = \lambda/(2\pir\epsilon0) is only valid for an infinitely long uniformly charged wire at a distance r from it, so it cannot be used to find E at point A. Therefore, simply plugging in 1m for x and setting it equal to 37 does not yield the correct value for k. Finally, since the electric
  • #1
n0va
7
0

Homework Statement


Thin rod AB has length L=100 cm and total charge q0=37 nC that is distributed in such a way that its line density [itex]\lambda[/itex] is proportional to the square of the distance from the end A, i.e. [itex]\lambda[/itex](x) =kx^2. Determine electric field E at the end A of the rod.

Homework Equations



E = (1/(4pi[itex]\epsilon[/itex]0))
Electric Field for a thin uniformly charged conducting wire: E = [itex]\lambda[/itex]/(2[itex]\pi[/itex]r[itex]\epsilon[/itex]0)

The Attempt at a Solution



if [itex]\lambda[/itex](x) =kx^2 , can we find k by plugging in 1m in x and setting it equal to 37. meaning k is 37.

Since they are asking for the E at point A, is x just 0? and does that mean that E is 0?
 
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  • #2
Does my answer make any sense at all? Or am i misinterpreting it? They're asking for the E field on point A, and since the distribution starts at A, x at that point is 0 and therefore there is no electricfield at point A?
 
  • #3
Cmon seriously?
 
  • #4
Hello n0va.

Welcome to Physics Forums (PF).

Did you read the https://www.physicsforums.com/showthread.php?t=94379", particularly #3 in this case.

To find k, you must integrate over the length of the charged rod & solve for k.[tex]Q=\int_{0m}^{1m}{kx^2}\,dx[/tex]

The formula you used to find E gives the E field at a distance, r, from an infinitely long charged rod which has uniform linear charge density.

To do this problem, you will have to do an integration over the length of the rod.
 
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  • #5
I suggest you to reread your posts before you send them and check the validity of your statements.

n0va said:

Homework Equations



E = (1/(4pi[itex]\epsilon[/itex]0))

What is E? If it denotes the electric field strength how can it be the same for all situations?

n0va said:
Electric Field for a thin uniformly charged conducting wire: E = [itex]\lambda[/itex]/(2[itex]\pi[/itex]r[itex]\epsilon[/itex]0)

This formula is valid for an infinite uniformly charged wire, at a distance r from it. The wire in the problem is neither infinite nor uniformly charged and the field is asked at zero distance from it, at one end.

n0va said:

The Attempt at a Solution



if [itex]\lambda[/itex](x) =kx^2 , can we find k by plugging in 1m in x and setting it equal to 37. meaning k is 37.

No, it is not right. The total charge is given. The integral of the charge density along the wire length is equal to the total charge, 37 nC, as SammyS suggested.

n0va said:
Since they are asking for the E at point A, is x just 0? and does that mean that E is 0?

The electric field has some contribution for all parts of the wire.

ehild
 

Related to Electric Field on finite charged rod

What is an electric field on a finite charged rod?

An electric field on a finite charged rod is a region in space where electrically charged particles experience a force. This force is exerted by the charged rod due to its electric charges.

How is the electric field strength calculated for a finite charged rod?

The electric field strength for a finite charged rod can be calculated using the formula E = kλ / r, where k is the Coulomb constant, λ is the linear charge density of the rod, and r is the distance from the rod to the point where the field is being measured.

What is the direction of the electric field on a finite charged rod?

The direction of the electric field on a finite charged rod is perpendicular to the rod and points away from the rod if the charge is positive, and towards the rod if the charge is negative.

How does the electric field change as you move along a finite charged rod?

The electric field strength decreases as you move further away from the charged rod. The direction of the electric field also changes depending on whether you are moving towards or away from the rod.

What are some real-world applications of the electric field on a finite charged rod?

The electric field on a finite charged rod is used in devices such as capacitors, which store electric energy, and in electrostatic precipitators, which remove particles from air. It is also used in electrostatic spraying, inkjet printing, and electrostatic painting techniques.

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