I really have trouble with solving this, me

  • Thread starter desparate
  • Start date
In summary: Yes, choose a point A at an angle θ from the middle, and a point B very close to A, at an angle θ + dθ, where dθ is very small. Then the charge of the section AB will be (dθ/π)Q,and since dθ is so small, we can assume that it is a point charge … that is, that the whole of AB is at the same distance (AP) from P. So the field will be of strength (dθ/π)Q/(AP)2, along the direction of AP. Carry on from there.
  • #1
desparate
4
0
http://img194.imageshack.us/img194/1559/45954311.png

Positive electric charge Q, total, are distributed uniformly on a semicircle
with radius a. What is the electric field (in moderation and direction) in the center
curvature (ie, point P)?

please help me, its the last i have to solve in order to finish my assignment, I am really desperate

Ps: I am from egypt, sorry for my english :(

ps1: this is from the book "university physics by D. Young"

thanks everyone in advance
 
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  • #2
Welcome to PF!

Hi desparate! Welcome to PF! :smile:

(btw, it's modulus! :wink:)

Find the field at P due to the charge between θ and θ + dθ, and integrate over -π/2 ≤ θ ≤ π/2. :smile:
 
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  • #3


tiny-tim said:
Hi desparate! Welcome to PF! :smile:

(btw, it's modulus! :wink:)

Find the field at P due to the charge between θ and θ + dθ, and integrate over -π/2 ≤ θ ≤ π/2. :smile:

can you tell me what the direction of the electric field is in P?
 
  • #4
desparate said:
can you tell me what the direction of the electric field is in P?

Hi desparate! :smile:

I've just noticed that the diagram doesn't match your question …

you asked for the field at the centre of curvature, which is O, not P. :confused:

The field at O seems a far more likely question (and symmetry will give you the direction there). :wink:

Before we go any further, can you please check which point is intended? :smile:
 
  • #5
doesnt it say P? (ie, P)?

so i think it has to be P
 
  • #6
Well, I'm highly doubtful.

(btw, is the book by Hugh D. Young?)

ok … then the field at P due to the charge between θ and θ + dθ will be in the direction of the chord from the point at θ to P.

You'll need to split it into x and y components before integrating. :smile:
 
  • #7
tiny-tim said:
Well, I'm highly doubtful.

(btw, is the book by Hugh D. Young?)

ok … then the field at P due to the charge between θ and θ + dθ will be in the direction of the chord from the point at θ to P.

You'll need to split it into x and y components before integrating. :smile:

yes its from this book, i really don't know why we are doing this kind of physics I am not studying physics or something like that

can you please draw something? i m not good at english, and i can't really understand what you mean by saying "the field at P due to the charge between θ and θ + dθ will be in the direction of the chord from the point at θ to P."

thanks for your help
 
  • #8
desparate said:
can you please draw something? i m not good at english, and i can't really understand what you mean by saying "the field at P due to the charge between θ and θ + dθ will be in the direction of the chord from the point at θ to P."

Yes, choose a point A at an angle θ from the middle, and a point B very close to A, at an angle θ + dθ, where dθ is very small.

Then the charge of the section AB will be (dθ/π)Q,

and since dθ is so small, we can assume that it is a point charge … that is, that the whole of AB is at the same distance (AP) from P.

So the field will be of strength (dθ/π)Q/(AP)2, along the direction of AP.

Carry on from there. :smile:
 

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