Again an electric potential question ?

In summary, the conversation is discussing a homework problem involving the calculation of electric potential at a point on the x-axis with a uniformly distributed charge. The conversation includes various equations and attempts at solving the problem, with the final solution involving the use of an improper integral to calculate the electric field along the x-axis. The conversation ends with the confirmation that the final answer is correct.
  • #1
Fazza3_uae
51
0
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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  • #2


how did you get this answer?
 
  • #3


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
 
  • #4


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.
 
  • #5


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 :)
 
  • #6


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]
 
  • #7


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 !
 
  • #8


Wow nice Job Friend , its absolutly right ^^

thank U very much ^^
 

1. What is electric potential?

Electric potential is the amount of electric potential energy per unit charge at a specific point in an electric field. It is measured in volts (V).

2. How is electric potential different from electric potential energy?

Electric potential is a measure of the potential energy per unit charge at a specific point, while electric potential energy is the total amount of potential energy possessed by an object due to its position in an electric field.

3. What is the equation for calculating electric potential?

The equation for electric potential is V = U/q, where V is electric potential (in volts), U is electric potential energy (in joules), and q is charge (in coulombs).

4. How is electric potential related to electric field?

Electric potential is directly related to electric field. The electric field is the force per unit charge acting on a charged object, while electric potential is the amount of potential energy per unit charge at a specific point in the electric field. They are related by the equation E = -ΔV/Δx, where E is electric field (in newtons per coulomb), V is electric potential (in volts), and x is the distance between two points (in meters).

5. How does electric potential affect the movement of charges?

Charges move from higher to lower electric potential, similar to how objects move from higher to lower gravitational potential. This movement is driven by the force exerted by the electric field. Positive charges move towards lower electric potential, while negative charges move towards higher electric potential.

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