Electric Field on finite charged rod

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n0va
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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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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?
 
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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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