Electric Field on finite charged rod

AI Thread Summary
To determine the electric field at point A of a finite charged rod with a non-uniform charge distribution, one must integrate the charge density along the rod's length rather than assuming a uniform field. The charge density is given by λ(x) = kx², and k must be calculated by integrating this expression over the rod's length to equal the total charge of 37 nC. The formula for electric field strength used in the discussion is applicable only to infinite uniformly charged wires, which does not apply here. At point A, the electric field is not zero, as contributions from all segments of the rod must be considered. Proper integration and understanding of the charge distribution are essential for finding the correct electric field value.
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 \lambda is proportional to the square of the distance from the end A, i.e. \lambda(x) =kx^2. Determine electric field E at the end A of the rod.

Homework Equations



E = (1/(4pi\epsilon0))
Electric Field for a thin uniformly charged conducting wire: E = \lambda/(2\pir\epsilon0)

The Attempt at a Solution



if \lambda(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?
 
Cmon seriously?
 
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.Q=\int_{0m}^{1m}{kx^2}\,dx

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\epsilon0))

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 = \lambda/(2\pir\epsilon0)

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 \lambda(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
 
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