Electric Field of a quarter ring

[Zayan]

Homework Statement

When calculating electric field of a quarter ring, putting limits from 0 to π/2 gives field lambda/4πER but when I put the limits as -theta1 to theta 2 taking theta from the middle symmetric axis,it gives a different value= lambda/2√2πER. Why does changing limits change the field?

Relevant Equations

dE=Integral kdq/r²

 

[kuruman]

Define your starting and ending angles conventionally counterclockwise with respect to the positive x-axis as shown in the figure on the right.
Then $$\begin{align} & dE_x=\frac{k\lambda}{R}\cos\theta~d\theta \nonumber \\
& dE_y=\frac{k\lambda }{R}\sin\theta ~d\theta.\nonumber \end{align}$$ Integrate and set the limiting angles $\theta_1$ and $\theta_2$ to whatever you like. One expression does all cases.

For example, in your case I, $\theta_1=\dfrac{\pi}{2}$ and $\theta_2=\pi.$

 

[TSny]

Why does changing limits change the field?

It's not surprising that $E_x$ is different for your two methods. Likewise for $E_y$.
But your two methods give the same result for the magnitude and direction of the net electric field vector.

 

[Zayan]

View attachment 347300Define your starting and ending angles conventionally with respect to the positive x-axis as shown in the figure on the right.
Then $$\begin{align} & dE_x=\frac{k\lambda}{R}\cos\theta~d\theta \nonumber \\
& dE_y=\frac{k\lambda }{R}\sin\theta ~d\theta.\nonumber \end{align}$$ Integrate and set the limiting angles $\theta_1$ and $\theta_2$ to whatever you like. One expression does all cases.

For example, in your case I, $\theta_1=\dfrac{\pi}{2}$ and ##\theta_2=\p

It's not surprising that $E_x$ is different for your two methods. Likewise for $E_y$.
But your two methods give the same result for the magnitude and direction of the net electric field vector.

Yessir it was the Orientation problem