# Electric Field of extended mass

by Redoctober
Tags: electric, extended, field, mass
 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 Attached Thumbnails
HW Helper
Thanks
PF Gold
P: 26,117
Hi Redoctober!
 Quote by Redoctober Integrating from 0 to pi ( thus taking only half of the semicircle ) …
noooo
0 to π/2
P: 41
 Quote by tiny-tim Hi Redoctober! noooo … 0 to π/2
Oh !! srrry typo error :S :S , i integrated from 0 to pi/2 .
The answer 2413 N/C is wrong :/ !

P: 41

## Electric Field of extended mass

Anyone has a solution ?? :/ !
 P: 389 The E field for the right part of your semi circle is given by $$E = \int_0^{\frac{\pi}{2}} \frac{-Q}{4\pi\epsilon_0r^2} d\theta$$ While the E field for the left part is the same but with +Q charge and integrated from $\frac{\pi}{2}$ to $\pi$. Just add these two together to get the total field.
 Sci Advisor Thanks P: 2,096 Are you sure? An electric field is a vector, and the correct formula is $$\vec{E}(\vec{x})=\frac{1}{4 \pi \epsilon_0} \int_{\mathbb{R}^3} \rho(\vec{x}') \frac{\vec{x}-\vec{x}'}{|\vec{x}-\vec{x}'|^3}.$$ 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.
 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 :/ !!
 P: 41 The final equation i got was actually correct i made an error with the calculation thats why my answer was wrong :)
 Sci Advisor Thanks P: 2,096 Yes, I've checked your result too. It's correct. I parametrized the line charge with help of the charge per angle: $$\lambda(\varphi)=\begin{cases} -\frac{2 Q}{\pi} & \text{for} \quad 0 \leq \varphi \leq \pi/2 \\ +\frac{2Q}{\pi} & \text{for} \quad \pi/2<\varphi\leq \pi \\ 0 & \text{elsewhere}. \end{cases}$$ Then you can use my formula for $\vec{x}=0$ to get $$\vec{E}=\frac{1}{4 \pi \epsilon_0 r^2} \int_0^{2 \pi} \mathrm{d} \varphi' \lambda(\varphi') \begin{pmatrix} -\cos \varphi' \\ -\sin \varphi' \\0 \end{pmatrix}.$$ Integrating over the two regions gives finally $$\vec{E}=\frac{Q}{\pi^2 \epsilon_0 r^2} \vec{e}_x.$$

 Related Discussions Introductory Physics Homework 3 Introductory Physics Homework 4 Introductory Physics Homework 1 Special & General Relativity 5 Classical Physics 0