Question

A plastic rod with linear charge density $\lambda$is bent into the quarter circle shown in FIGURE. We want to find the electric field at the origin.

a. Write expressions for the $x$- and $y$-components of the electric field at the origin due to a small piece of charge at angle $\theta$.

b. Write, but do not evaluate, definite integrals for the $x$- and $y$-components of the net electric field at the origin.

c. Evaluate the integrals and write ${\overrightarrow{E}}_{\text{net}}$in component form

Short answer

(a) Expression for components of field,$d{E}_x=\frac{k\lambda }{R}$and$d{E}_x=\frac{k\lambda }{R}\sin \theta \Delta \theta$

(b) Electrical field at the origin, $E_x=\frac{k\lambda }{R}{\int }_0^{\pi /2}\cos \theta \Delta \theta$and $E_y=\frac{k\lambda }{R}{\int }_0^{\pi /2}\sin \theta \Delta \theta$

(c)$\overrightarrow{E}$in component form,$\overrightarrow{E}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q}{\pi R^2}\right)(\hat{i}+\hat{j})$

Step-by-step solution

Find Expressions for component of field (part a)

The force experienced by such a unit positive charge put at a spot is the electric field intensity at that point. The intensity of an electric field is a vector quantity.

$E=\frac{kq}{R^2}$

where $R$denotes distance, $k$denotes Coulomb's constant, and $q$denotes charge

$dE=\frac{k\Delta q}{R^2}$

and

$x=R\theta ,\Delta x=R\Delta \theta$

(From figure)

$q=\lambda x,\Delta q=\lambda \left(\Delta x\right)$

(charge in terms of density)

$⟶\Delta x=R\Delta \theta$and $\Delta q=\lambda \left(R\Delta \theta \right)$

Fill in the blanks in the equation with these values.,

$dE=\frac{k\lambda \Delta \theta }{R}$

As a result, the electric field's $x$component will be,

$$\begin{gathered} d{E}_x=dE\cos \theta \\ =\frac{k\lambda \Delta \theta }{R}\cos \theta \end{gathered}$$

$\to d{E}_x=\frac{k\lambda }{R}\cos \theta \Delta \theta$

As a result, the electric field's $y$component will be,

$\to d{E}_y=\frac{k\lambda }{R}\sin \theta \Delta \theta$

Find Electrical field at the origin (part b)

$x$component equation,

$d{E}_x=\frac{k\lambda }{R}\cos \theta \Delta \theta$

Integrate this equation,

$E_x=\frac{k\lambda }{R}{\int }_0^{\pi /2}\cos \theta \Delta \theta$ $\left(1\right)$

$y$component equation,

$d{E}_y=\frac{k\lambda }{R}\sin \theta \Delta \theta$

Integration of this equation,

$E_y=\frac{k\lambda }{R}{\int }_0^{\pi /2}\sin \theta \Delta \theta$ $\left(2\right)$

Linear charge density of the rod,

$\lambda =\frac{2Q}{\pi R}$

substitute this value in equation $\left(1\right)$and $\left(2\right)$

$E_x=\frac{k2Q}{\pi R^2}{\int }_0^{\pi /2}\cos \theta \Delta \theta$

$E_y=\frac{k2Q}{\pi R^2}{\int }_0^{\pi /2}\sin \theta \Delta \theta$

Calculate the integral using both equations,

${\int }_0^{\pi /2}\cos \theta d\theta =[\sin \theta {]}_0^{\pi /2}=+1$

${\int }_0^{\pi /2}\sin \theta d\theta =[-\cos \theta {]}_0^{\pi /2}=+1$

Find electric field in component form

the electric field can be written as:

$\overrightarrow{E_{net}}=E_x\hat{i}+E_y\hat{j}$

Expand:

$E_x\hat{i}=\left[\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q}{\pi R^2}\right){\int }_0^{\pi /2}\cos \theta \Delta \theta \right]\hat{i}$

$E_y\hat{j}=E_x\hat{i}=\left[\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q}{\pi R^2}\right){\int }_0^{\pi /2}\sin \theta \Delta \theta \right]\hat{j}$

the sum of equations will be,

localid="1650218257527" $\overrightarrow{E_{net}}=\frac{1}{4\pi {\epsilon }_0}\left(\frac{2Q}{\pi R^2}\right)(\hat{i}+\hat{j})$