Electric Field of extended mass

In summary, Vanhees looks for the electric field in the direction of the x-axis at the origin of a semicircle of charge +Q and charge -Q. He finds that the electric field is 2413 N/C.
  • #1
Redoctober
48
1
So here is the scenario (see attachment) - I have a semicircle wire (radius R=15.9cm) which is made of insulator material , the semicircle consist of two combined quartercircle wires parts where one has equally distributed charge +Q and the other has -Q . Required is find the Electric field in direction of x at the origin . Q=5.33nC

My approach was as follows

Let E = 1/(4*pi*e)∫1/(R^2).dQ r

dQ=λ*ds and ds=R*dθ and i also know that unit vector r = cosθ*i+sinθ*j

therefore for the E in x direction i get this expression

E = 1/(4*pi*e)*1/(R^2)*λ*R∫cosθ.dθ

Integrating from 0 to pi ( thus taking only half of the semicircle ) and using λ as 2/(pi*r)

I get Q/(2*pi^2*e*R^2) .
Because the other half has opposite charge i can say that the Etot = Eneg +Epos

Therefore i multiply the equation by two to finaly get

Q/(pi^2*e*R^2)

If i put the values given i get as absolute value 2413 N/C for Electric field at origin of circel in the direction of x

Unfortunately it is a wrong solution :( ! What is the mistake i hv done ?? Can anyone spot it ? Thanks in advance
 

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  • #2
Hi Redoctober! :smile:
Redoctober said:
Integrating from 0 to pi ( thus taking only half of the semicircle ) …

noooo :cry:

0 to π/2 :redface:
 
  • #3
tiny-tim said:
Hi Redoctober! :smile:


noooo :cry:

0 to π/2 :redface:

Oh ! srrry typo error :S :S , i integrated from 0 to pi/2 .
The answer 2413 N/C is wrong :/ !
 
  • #4
Anyone has a solution ?? :/ !
 
  • #5
The E field for the right part of your semi circle is given by
[tex]E = \int_0^{\frac{\pi}{2}} \frac{-Q}{4\pi\epsilon_0r^2} d\theta[/tex]
While the E field for the left part is the same but with +Q charge and integrated from [itex]\frac{\pi}{2}[/itex] to [itex]\pi[/itex]. Just add these two together to get the total field.
 
  • #6
Are you sure? An electric field is a vector, and the correct formula is

[tex]\vec{E}(\vec{x})=\frac{1}{4 \pi \epsilon_0} \int_{\mathbb{R}^3} \rho(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.[/tex]

Of course, in the here considered case, you have to integrate along the semicircle and use the charge per length instead of the bulk-charge density. However, you should check, whether you have all the geometrical factors for you vector component right.
 
  • #7
Vanhees is right , Electric field is a vector .
So regarding the E will be in radial direction .
For the Y axis we need to consider the j component .

Vanhees , your mathematical expression is a bit high beyond my math skill xD !

My question is why the final expression for the problem is wrong :/

i considered the +Q and -Q E vector addition

I reached to the conclusion of

|E| = Q/(4*pi^2*e*r^2) (In the direction of J

Btw : how do u write the math expressions in that style xD . my way of typing the math is silly :/ !
 
  • #8
The final equation i got was actually correct

i made an error with the calculation that's why my answer was wrong :)
 
Last edited:
  • #9
Yes, I've checked your result too. It's correct. I parametrized the line charge with help of the charge per angle:

[tex]\lambda(\varphi)=\begin{cases}
-\frac{2 Q}{\pi} & \text{for} \quad 0 \leq \varphi \leq \pi/2 \\
+\frac{2Q}{\pi} & \text{for} \quad \pi/2<\varphi\leq \pi \\
0 & \text{elsewhere}.
\end{cases}
[/tex]

Then you can use my formula for [itex]\vec{x}=0[/itex] to get

[tex]\vec{E}=\frac{1}{4 \pi \epsilon_0 r^2} \int_0^{2 \pi} \mathrm{d} \varphi' \lambda(\varphi') \begin{pmatrix} -\cos \varphi' \\ -\sin \varphi' \\0 \end{pmatrix}.[/tex]

Integrating over the two regions gives finally

[tex]\vec{E}=\frac{Q}{\pi^2 \epsilon_0 r^2} \vec{e}_x.[/tex]
 

What is an electric field of extended mass?

The electric field of extended mass refers to the area surrounding a large object that has a net electric charge. It is a measure of the force that would be exerted on a charged particle placed at any point within the field.

How is the electric field of extended mass calculated?

The electric field of extended mass is calculated using Coulomb's Law, which states that the electric field strength is inversely proportional to the square of the distance from the charged object and directly proportional to the magnitude of the charge.

What are some examples of objects with an electric field of extended mass?

Some examples of objects with an electric field of extended mass include planets, stars, and even everyday objects like a balloon or a comb that have a static charge.

How does the electric field of extended mass affect charged particles?

The electric field of extended mass can either attract or repel charged particles, depending on the direction of the field and the charge of the particle. Charged particles will experience a force in the direction of the electric field lines.

What are some real-life applications of the electric field of extended mass?

The electric field of extended mass has many practical applications, including the functioning of electronic devices, the movement of charged particles in a particle accelerator, and the study of atmospheric phenomena such as lightning and auroras.

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