Q71P - Fundamentals Of Physics

Question

A charge of  $\mathbf{20}\mathbf{ }\mathbf{\text{nC}}$is uniformly distributed along a straight rod of length $\mathbf{4}\mathbf{.0}\mathbf{ }\mathbf{\text{m}}$ that is bent into a circular arc with a radius of $\mathbf{2}\mathbf{.0}\mathbf{ }\mathbf{\text{m}}$ .What is the magnitude of the electric field at the center of curvature of the arc?

Step-by-Step Solution

Answer

The magnitude of the electric field at the centre of curvature of the arc is.$38 \text{N/C}$

 

Step 1: The given data

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,$R=2\text{m}$

Step 2: Understanding the concept of the electric field

Consider an infinitesimal section of the arc of length dx. It contains charge$dq=\lambda dx$ 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:                                                 $dE=\frac{1}{4\pi {\epsilon }_o}\frac{\lambda dx}{r^2}$                             (i)

Step 3: Calculation of the electric field at the center of the curvature of the arc

“Electric field of a charged circular rod,” 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:

$\overrightarrow{E}=\frac{\lambda sin\theta }{4\pi {ϵ}_{o}r}{|}_{-\theta }^{\theta }\mathrm{........................}\left(a\right)$

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

Thus, the angle is given as:

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

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

$\begin{array}{c}\left|\overrightarrow{E}\right|=\frac{(q/l) sin\theta }{4\pi {ϵ}_{o}r}{|}_{ -1.0 \text{rad} }^{1.0 \text{rad}}\\ =38 \text{N/C}\end{array}$

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