Magnitude of electric field at point A using components

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Homework Help Overview

The discussion revolves around calculating the magnitude of the electric field at a specific point due to a charged rod. The problem involves understanding the contributions of electric field components and integrating over a defined length of the rod.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the cancellation of y-components of the electric field and the need to focus on the x-components. There is an exploration of the integration process for calculating the electric field contributions from differential charge elements.

Discussion Status

Some participants have provided insights into the integration of electric field components and have raised questions about specific angles and the correct application of formulas. There is an ongoing exploration of the calculations leading to discrepancies between the participant's results and the expected answer.

Contextual Notes

Participants are working with a specific geometry and charge distribution, and there is mention of potential confusion regarding the angle used in calculations. The original poster expresses uncertainty about the integration limits and the relationship between differential charge and arc length.

Rijad Hadzic
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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?
 
Last edited:
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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:
stock-vector-vector-illustration-of-the-ancient-greek-alphabet-simple-pic-470453465.jpg
 
Can anyone help me with this one :< sorry to bump
 
Sorry to bump again. I know there is a lot to read and follow but if anyone can help me I would appreciate it
 
Rijad Hadzic said:

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 ∂?

Rijad Hadzic said:
- Electric field formula product cos(greekletter) to get the x component only
Which Greek letter?
Rijad Hadzic said:
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.
 

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