Potential of a circle boat

In summary, the conversation discusses the calculation of electric field and potential for a line charge distributed over a circle, with radius a and sector 2β. The formula for electric field is given, but there is a step missing to evaluate it. The potential can be calculated using the formula ∫ (λ(r')/R) dl', but it is much more work because there is no simple expression for E at other points. The conversation also addresses the inclusion of φ in the formula for electric field, which is incorrect as the electric field can only depend on parameters of the setup. The easiest way to find V is through the formula V(r) = (1/4πε) ∫ (λ(r')/R) dl'.
  • #1
Pouyan
103
8

Homework Statement


A line charge has the total charge Q evenly distributed over a circle boat with radius a and sector 2β, placed according to the figure

figure.jpg

ctUmHk


Find the Electric field E and the potential V in the origin.

Homework Equations


I know for this case that E(r) = (1/4πε) ∫ (λ(r')/R2)R dl' , R is unit a vector

λ(r') = Q/(2βa) R = - a *(cos(φ)x + sin(φ)y) dl'=a*dφ R2= a^2

V(r) = (1/4πε) ∫ (λ(r')/R) dl'

The Attempt at a Solution


The electric field is not so hard to calculate I know
E(r) =Q/(2βa)(1/4πε) ∫ ((-a *(cos(φ)x + sin(φ)y))/a3) adφ
which is : (Q/8πεa2β) * [ -sinφx + cosφy]

Now if I want to calculate the potential I know :
V(r) = ∫ E*dl where dl = a*dφ*R = -a*dφ*(cos(φ)x + sin(φ)y)
But I see (Q/8πεa2β) * [ -sinφx + cosφy] * (-a*dφ*(cos(φ)x + sin(φ)y)) = 0!

And the real solution is :

V(r) = (1/4πε) ∫ (λ(r')/R) dl' = (1/4πε) * (Q/2βa) ∫(adφ/a) and for a boat circle from π-β to π+β

Or :
sol-jpg.jpg


What is actually wrong ?! If I do the same algorithm for a Sphere, it gives me a correct answer but why can't I do the same thing here through ∫ E*dl ?
 
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  • #2
Pouyan said:
which is : (Q/8πεa2β) * [ -sinφx + cosφy].
There is a step missing to evaluate the electric field.

For the potential, there is an easier way, without using the electric field. Otherwise you would have to calculate the electric field everywhere along your integration path and that gets much more complicated.
 
  • #3
mfb said:
There is a step missing to evaluate the electric field.

For the potential, there is an easier way, without using the electric field. Otherwise you would have to calculate the electric field everywhere along your integration path and that gets much more complicated.
Ok the whole solution is :

sol.jpg
Why can't we just integrate E*dl from this solution for E ?!
 
  • #5
mfb said:
That looks right.
But is this wrong to think that E =(Q/8πεa2β) * [ -sinφx + cosφy] ?! If we just want to find V through E *dl ?!
I know that there is an easier way. Just using formula. But when I will find potential for a solid sphere it seems much more easier to use ∫E*dl. First we find E and then we integrate over that E and find V. But here I see we can not do exact the same thing ... why ?!
 
  • #6
What is φ in that formula? How can the electric field depend on it?
The electric field has a zero y component due to the symmetry of the setup.
Pouyan said:
But here I see we can not do exact the same thing ... why ?!
You can do it, but it is much more work because there is no simple expression for E at other places.
 
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  • #7
mfb said:
What is φ in that formula? How can the electric field depend on it?
The electric field has a zero y component due to the symmetry of the setup.
You can do it, but it is much more work because there is no simple expression for E at other places.
It's for polar coordinate.
 
  • #8
mfb said:
You can do it, but it is much more work because there is no simple expression for E at other places.
OK and because of that the easiest way is : V(r) = (1/4πε) ∫ (λ(r')/R) dl' ...
 
  • #9
Pouyan said:
But is this wrong to think that E =(Q/8πεa2β) * [ -sinφx + cosφy] ?! If we just want to find V through E *dl ?!
Yes, it's wrong. You're integrating over ##\phi## with definite limits, so that variable shouldn't appear in the final result.

I know that there is an easier way. Just using formula. But when I will find potential for a solid sphere it seems much more easier to use ∫E*dl. First we find E and then we integrate over that E and find V. But here I see we can not do exact the same thing ... why ?!
You're calculating the electric field at one point, the origin. To find the potential by integration, you need to find ##\vec{E}## at all points along whatever path you choose from infinity to the origin. In other words, what you're proposing to do is calculate
$$-\int_C \vec{E}(0)\cdot d\vec{r}$$ instead of
$$-\int_C \vec{E}(r)\cdot d\vec{r}$$ where ##C## is some path from infinity to the origin.
 
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  • #10
Pouyan said:
It's for polar coordinate.
The polar coordinate of what?
The electric field can only depend on parameters of the setup (##\beta##, a and Q). It cannot depend on intermediate parameters introduced in the calculation. If someone gives you values for these three things you have to be able to calculate the electric field in V/m. What would you plug in for ##\phi##?
Pouyan said:
OK and because of that the easiest way is : V(r) = (1/4πε) ∫ (λ(r')/R) dl' ...
Right.
 
  • #11
vela said:
Yes, it's wrong. You're integrating over ##\phi## with definite limits, so that variable shouldn't appear in the final result.You're calculating the electric field at one point, the origin. To find the potential by integration, you need to find ##\vec{E}## at all points along whatever path you choose from infinity to the origin. In other words, what you're proposing to do is calculate
$$-\int_C \vec{E}(0)\cdot d\vec{r}$$ instead of
$$-\int_C \vec{E}(r)\cdot d\vec{r}$$ where ##C## is some path from infinity to the origin.

OK than you !

What do I see; it's not easy to find E(r) for this case...
 

1. What is the potential of a circle boat?

The potential of a circle boat refers to its ability to move efficiently and smoothly through water. This is influenced by factors such as the boat's design, weight, and power source.

2. How is the potential of a circle boat measured?

The potential of a circle boat is typically measured by its speed, stability, and maneuverability. These factors can be assessed through various tests and simulations.

3. What are the advantages of a circle boat's potential?

A circle boat with high potential can offer benefits such as increased speed, improved fuel efficiency, and better handling in rough waters. This can be advantageous for activities such as fishing, recreational boating, and water sports.

4. Can the potential of a circle boat be improved?

Yes, the potential of a circle boat can be improved through various means such as optimizing its design, using lightweight materials, and upgrading its power source. Regular maintenance and upgrades can also help maintain the boat's potential over time.

5. Are there any limitations to a circle boat's potential?

Like any other vessel, a circle boat's potential may be limited by external factors such as weather conditions, water depth, and the skill of the operator. Additionally, the potential of a circle boat may vary depending on its intended use and design.

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