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Electric Field of extended mass 
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#1
Feb1612, 04:29 AM

P: 41

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 


#3
Feb1612, 06:26 AM

P: 41

The answer 2413 N/C is wrong :/ ! 


#4
Feb1612, 10:12 AM

P: 41

Electric Field of extended mass
Anyone has a solution ?? :/ !



#5
Feb1712, 01:52 AM

P: 389

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
Feb1712, 03:24 AM

Sci Advisor
Thanks
P: 2,486

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 bulkcharge density. However, you should check, whether you have all the geometrical factors for you vector component right. 


#7
Feb1712, 05:28 AM

P: 41

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
Feb1912, 02:51 AM

P: 41

The final equation i got was actually correct
i made an error with the calculation thats why my answer was wrong :) 


#9
Feb1912, 04:06 AM

Sci Advisor
Thanks
P: 2,486

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] 


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