# Electric field in a semi-circle (non-uniformly charged)

• BitterX
In summary, the homework problem involves finding the electric field at a point where the radius of a plastic semi-circle with non-uniformly distributed charge intersects. The charge density is given by lambda = lambda_0 sin(theta), with theta being 0 at the middle of the arc. The equations used for solving the problem are dE = (k*dQ)/R^2 and phi = integral of (E*dR). The solution involves separating the electric field into x and y components and integrating from 0 to pi for x and 0 to -pi for y, resulting in both components having the same magnitude pointing in opposite directions.
BitterX

## 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 $\lambda = \lambda_0 sin\theta$
θ is 0 in the middle of the arc

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

## Homework Equations

$dE=\frac{kdQ}{R^2}$
$\phi = \int{Edr}$

## The Attempt at a Solution

charge differential to angle differential is :
$dQ=\lambda R = \lambda_0 sin\theta d\theta$

for the field I separated for x and y
$E_x =\frac{k\lambda_0}{R} \int_{0}^{\pi}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{2R}$

the same way I get
$E_y= \frac{k\lambda_0}{R}\int_{0}^{\pi}sin\theta cos\theta d\theta=0$

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 $\frac{\pi}{2}$ and once from 0 to $-\frac{\pi}{2}$

Last edited:
from each one, so E_x =\frac{k\lambda_0}{R} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin^2\theta d\theta=\frac{k\lambda_0}{R}=\frac{k\lambda_0\pi}{R} and E_y= \frac{k\lambda_0}{R}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}sin\theta cos\theta d\theta=\frac{k\lambda_0\pi}{R} so the electric field is pointing in both directions with the same magnitude.

## What is an electric field in a semi-circle?

An electric field in a semi-circle refers to the distribution of electric charges within a semi-circular shape, which can be either uniformly or non-uniformly charged. It describes the direction and magnitude of the electric force that a charged particle would experience at any point within the semi-circle.

## How is the electric field calculated in a semi-circle?

The electric field in a semi-circle can be calculated using Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. By considering the distribution of charges within the semi-circle, the electric field can be calculated at any point using this law.

## What factors affect the electric field in a semi-circle?

The electric field in a semi-circle is affected by the magnitude and distribution of the charges within the semi-circle, as well as the distance from the charged particles. Additionally, the presence of other external charges or conductors nearby can also influence the electric field.

## How does the electric field vary in a non-uniformly charged semi-circle?

In a non-uniformly charged semi-circle, the electric field varies at different points due to the uneven distribution of charges. This means that the electric field will be stronger at certain points and weaker at others, depending on the location of the charged particles within the semi-circle.

## What is the significance of studying the electric field in a semi-circle?

Studying the electric field in a semi-circle is important in understanding the behavior of electric charges and their interactions. This knowledge has practical applications in various fields, such as electronics, engineering, and physics, and can also help in predicting and controlling the movement of charged particles in a given system.

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