- #1

BitterX

- 36

- 0

## Homework Statement

A plastic semi-circle/arc with radius

**R**has a non-uniformly distributed charge upon it.

The density of the charge/length is [itex]\lambda = \lambda_0 sin\theta[/itex]

**θ**is 0 in the middle of the arc

find the electric field in the point where the radius come from

## Homework Equations

[itex]dE=\frac{kdQ}{R^2}[/itex]

[itex]\phi = \int{Edr}[/itex]

## The Attempt at a Solution

charge differential to angle differential is :

[itex]dQ=\lambda R = \lambda_0 sin\theta d\theta[/itex]

for the field I separated for x and y

[itex]E_x =\frac{k\lambda_0}{R} \int_{0}^{\pi}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{2R} [/itex]

the same way I get

[itex]E_y= \frac{k\lambda_0}{R}\int_{0}^{\pi}sin\theta cos\theta d\theta=0 [/itex]

this is counter-intuitive for me, I thought the electric field should be pointing in the y direction.

I guess I need to have my answers inverted, but would be glad to understand why, and maybe shed some light on my mistakes...edit: I think I've found my mistake... I should integrate once from 0 to [itex]\frac{\pi}{2}[/itex] and once from 0 to [itex]-\frac{\pi}{2}[/itex]

and add what I get

Last edited: