Again an electric potential question ?

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Homework Help Overview

The discussion revolves around calculating the electric potential due to a uniformly distributed charge along the x-axis, specifically for a point located at 5 m. The original poster expresses confusion regarding their initial calculation of the potential and seeks clarification on the correct approach.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the formula for electric potential and the necessity of integrating the electric field along a path to compute the potential correctly. Questions arise about the validity of using certain equations and the interpretation of distance in the context of the problem.

Discussion Status

Some participants have provided clarifications regarding the integration process needed to find the potential, while others express uncertainty about the implications of integrating to infinity and the correct limits for the integration. There is an acknowledgment of the need for a proper setup to avoid incorrect conclusions.

Contextual Notes

Participants are navigating the complexities of improper integrals and the specific setup required for this problem, indicating a need for careful consideration of assumptions and definitions related to electric potential.

Fazza3_uae
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Again an electric potential question ?

Homework Statement




Question # 1

A charge of 7.034 nC is uniformly distributed
along the x-axis from −4 m to 4 m.
What is the electric potential (relative to
zero at infinity) of the point at 5 m on the
x-axis?





Homework Equations





The Attempt at a Solution




I got 21.0728 and its wrong :cry: How come ??
 
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how did you get this answer?
 


The Attempt at a Solution



I know that E in this case = kq/[a*(L+a)]

a : distance from point P to x2 ( P-x2)
L : distance from x1 to x2 ( x2-x1)

The denominator = r²= [a*(L+a)]

I want Potential V=kq/r

r=Sqr(The denominator = r²= [a*(L+a)])

So I just insert the values and got that answer *_* which is wrong
 


Fazza3_uae said:
I want Potential V=kq/r


V = kq/r is only valid for a point charge or a uniformly charged sphere.
You have to integrate [itex]E \bullet ds[/itex] along a path from the
point with x = 5 to infinity.
 


thx 4 clarifying about the potential ^^

But is'nt taking an integral from 5 to infinity will give me infinity and when i multiply infinity with E i will get also infinity ?


In case what u r saying is correct what is the value of r then ??

Is it the distance from point p to x1 or to x2 ? or its something else !

Appreciate Ur Help in Advance Mr. willem2 :)
 


This computation doesn't have an r in it. you already gave an expression for E without
R in it. you can integrate the field along the x axis, so E has the same direction as ds, so you
get

[tex]\int_5^{\infty} \frac {k q} {(x-x_1)(x-x_2)}dx[/tex]


This kind of integral with infinity as a limit is called an improper integral and it is actually a
limit:

[tex]\int_a^{\infty} f(x) dx = \lim_{L \to +\infty} \int_a^L f(x) dx[/tex]
 


Aha , i get it now. I will submitt the answer & i am sure it will be correct this time , I'll be right back in minutes ^^

THx again Boss ^^ Really appreciate Ur help !
 


Wow nice Job Friend , its absolutly right ^^

thank U very much ^^
 

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