- #1

Zack K

- 166

- 6

## Homework Statement

A rod of charged -Q is curved from the x-axis to angle ##\alpha##. The rod is a distance R from the origin (I will have a picture uploaded). What is the electric field of the charge in terms of it's x and y components at the origin? k is ##\frac {1} {4\pi \epsilon_0}##

## Homework Equations

##\vec E=\frac {kQ} {r^2}##

## The Attempt at a Solution

Let ##\theta## represent a any angle of ##\alpha##

##\vec r=\langle -Rcos\theta, -Rsin\theta, 0\rangle##

##|\vec r|=\sqrt {(-Rcos\theta)^2+(-Rsin\theta)^2}##

$$\hat r=\frac {\vec r} {\vec |r|}=\frac {\langle-cos\theta, -sin\theta, 0\rangle} {\sqrt{-cos\theta)^2+(-sin\theta)^2}}=\frac {\langle-cos\theta, -sin\theta, 0\rangle}{1}$$You can factor out the R from the top and bottom and cancel them out. The denominator just becomes 1

Now representing ##\Delta Q##

$$\Delta Q=Q\frac{\Delta arclength} {arclength}=Q\frac{R\Delta\theta} {R\alpha}=Q\frac{\Delta\theta} {\alpha}$$

The R's will cancel out.

Now assuming rod consists of many point charges

$$\Delta {\vec E}=\frac {k\Delta Q}{|\vec r|^2} \hat r=\frac {kQ\Delta \theta}{R^2\alpha}\langle-cos\theta, -sin\theta, 0\rangle$$Now we integrate in terms of ##\theta## and integrate each component. $$\Delta {\vec E_x}=\int_{0}^{\alpha} \frac {kQ(-cos\theta)}{R^2\alpha}d\theta=\frac {kQ}{R^2}\int_{0}^{\alpha} \frac {(-cos\theta)}{\alpha}d\theta$$ $$\Delta {\vec E_y}=\int_{0}^{\alpha} \frac {kQ(-sin\theta)}{R^2\alpha}d\theta=\frac {kQ}{R^2}\int_{0}^{\alpha} \frac {(-sin\theta)}{\alpha}d\theta$$This is where I'm stuck on. Not entirely sure if I can factor an alpha out.