91影视

1. Charge that is uniformly distributed,$q=20\text{nC}$
2. Length of the rod,$l=2\text{m}$
3. Radius of the circular arc,$R=2\text{m}$

Consider an infinitesimal section of the arc of length dx. It contains charge$dq=位dx$and is a distance r from the center. Thus, using this concept of the electric field we can get the required value of the field at the center of the arc.

Formula:

The magnitude of the field due to this element at the centre is given by: $dE=\frac{1}{4蟺{蔚}_o}\frac{位dx}{r^2}$ (i)

鈥淓lectric field of a charged circular rod,鈥� we see that the field evaluated at the center of curvature due to a charged distribution on a circular arc is given by equation (i) as follows:

$\stackrel{鈫�}{E}=\frac{位sin胃}{4蟺{系}_{o}r}{|}_{鈭�胃}^{胃}\mathrm{........................}\left(a\right)$

Along the symmetry axis, where,$位=q/l\text{or}q/r胃$, $胃$in radians.

Thus, the angle is given as:

$\begin{array}{c}胃=l/r\\ =4.0/2.0\\ =2.0\text{rad}\end{array}$

Now, with, $q=20x{10}^{鈥�9}\text{C}$we obtain the electric field using equation (a) as:

$\begin{array}{c}\left|\stackrel{鈫�}{E}\right|=\frac{(q/l)sin胃}{4蟺{系}_{o}r}{|}_{鈭�1.0\text{rad}}^{1.0\text{rad}}\\ =38\text{N/C}\end{array}$

Hence, the value of the electric field is.$38\text{N/C}$