Magnitude of electric field at point A using components

[Rijad Hadzic]

Homework Statement

https://imgur.com/gallery/YwCap

In side are image of figure, and questions.

Homework Equations

E = kq/r^2

The Attempt at a Solution

So I have an electric field where the y compontents cancel.

I have $2Ecos(∂) = E (Electric field at point A)$

- 2 because there is a top and bottom part of rod with same magnitude.

- Electric field formula product cos(greekletter) to get the x component only

I have $E = kQ/R^2$

big R because that's the radius given, and it will never change.

$Q/L = \Lambda, dQ = dL\Lambda$

Okay now here is where I'm confused...

I can set $dL = R*d(∂)$, right? Because essentially that is the arclength, right?!?

$dQ = R*d(∂) * \Lambda$

so $dE =( R*d(∂) * \Lambda * k ) / R^2$

pull out constants

E = (R * lambda * k ) / R^2 * integral of d(∂), from 0 to R∂, (the acrlength of the rod)

So now you have

$E = R^2 * \Lambda * k * ∂ / R^2$

but $Q = \Lambda * ∂ * R$

$E = Q*R*k / R^2$

$E = Q * k / R$

but

$2Ecos(∂) = E$

so I have

2*Q * k * cos(∂) / R

Now plugging in the values given q = 35.5 x 10^-9 R = .785 m ∂ = 60 degrees k = 8.99x10^9

I get answer 407 N/C but my book is telling me 428. Can anyone explain what I did wrong?

 

[ehild]

When writing a post, hit the Σ key, it provides Greek letters. Clicking on the desired one, it appears in your text. By the way, this is the Greek alphabet:

 

[Rijad Hadzic]

Can anyone help me with this one :< sorry to bump

 

[Rijad Hadzic]

Sorry to bump again. I know there is a lot to read and follow but if anyone can help me I would appreciate it

 

[ehild]

Homework Statement

https://imgur.com/gallery/YwCap

In side are image of figure, and questions.

Homework Equations

E = kq/r^2

The Attempt at a Solution

So I have an electric field where the y compontents cancel.

I have $2Ecos(∂) = E (Electric field at point A)$

- 2 because there is a top and bottom part of rod with same magnitude.

What is the angle ∂?

- Electric field formula product cos(greekletter) to get the x component only

Which Greek letter?

I have $E = kQ/R^2$

big R because that's the radius given, and it will never change.

$Q/L = \Lambda, dQ = dL\Lambda$

Okay now here is where I'm confused...

I can set $dL = R*d(∂)$, right? Because essentially that is the arclength, right?!?

$dQ = R*d(∂) * \Lambda$

so $dE =( R*d(∂) * \Lambda * k ) / R^2$

pull out constants

E = (R * lambda * k ) / R^2 * integral of d(∂), from 0 to R∂, (the acrlength of the rod)

So now you have

$E = R^2 * \Lambda * k * ∂ / R^2$

but $Q = \Lambda * ∂ * R$

$E = Q*R*k / R^2$

$E = Q * k / R$

but

$2Ecos(∂) = E$

so I have

2*Q * k * cos(∂) / R

Now plugging in the values given q = 35.5 x 10^-9 R = .785 m ∂ = 60 degrees k = 8.99x10^9

I get answer 407 N/C but my book is telling me 428. Can anyone explain what I did wrong?

You have to add up (integrate) the x components of the electric fields due to the charge of a line element dΛ=Rd∂, between 0 and Φ.