Question

A thread charged homogeneously with line density λ=3.47 μC/m forms a loop in the shape of a cut circle with radius r=6.12m, cut by a line parallel to the y-axis with length l=10.54. Calculate the elctric field at the center of the circle. State direction and magnitude of the electric field vector.

Answer 1

Charge line density $\lambda = 3.47 \, \mu\text{C/m}$

Radius of circle $r = 6.12 \, \text{m}$

Length of cut $L = 10.54 \, \text{m}$

Lets divide arc by small parts with angle $d\varphi$ . It has length $dl = rd\varphi$ and charge $dq = \lambda r d\varphi$ , therefore produces electric field with magnitude

$$dE_y = - \frac{1}{4\pi\varepsilon_0}\frac{\lambda r d \varphi}{r^2} \sin \varphi, \, dE_y = - \frac{1}{4\pi\varepsilon_0}\frac{\lambda r d \varphi}{r^2} \cos \varphi$$

As electromagnetic field obeys superposition principle, therefore we integrate in ranges from $-\theta$ to $\theta$ :

$$E_x = - \frac{1}{4\pi\varepsilon_0}\int_{-\theta}^{\theta} \frac{\lambda r d \varphi}{r^2} \sin \varphi = \frac{\lambda}{4\pi\varepsilon_0 r} \cos \varphi |_{-\theta}^{\theta} = 0$$

$$E_y = - \frac{1}{4\pi\varepsilon_0}\int_{-\theta}^{\theta} \frac{\lambda r d \varphi}{r^2} \cos \varphi =- \frac{\lambda}{4\pi\varepsilon_0 r} \sin \varphi |_{-\theta}^{\theta} = - \frac{\lambda}{2\pi\varepsilon_0 r} \sin \theta$$

where θ defines as

$$sin \theta = \frac{L}{2r}$$

Therefore, magnitude of electric field equals to abs. value of y component and directed parallel to y-line opposite to arc.

$$E_y = - \frac{\lambda L}{4\pi\varepsilon_0 r^2},\, E_x=0$$

$$E=\frac{\lambda L}{4\pi\varepsilon_0 r^2}= 8788 \text{V/m}$$