1. Oct 20, 2009

### desparate

http://img194.imageshack.us/img194/1559/45954311.png [Broken]

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, im really desperate

Ps: im from egypt, sorry for my english :(

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

Last edited by a moderator: May 4, 2017
2. Oct 20, 2009

### tiny-tim

Welcome to PF!

Hi desparate! Welcome to PF!

(btw, it's modulus! )

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

Last edited: Oct 20, 2009
3. Oct 21, 2009

### desparate

Re: Welcome to PF!

can you tell me what the direction of the electric field is in P?

4. Oct 21, 2009

### tiny-tim

Hi desparate!

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.

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

Before we go any further, can you please check which point is intended?

5. Oct 21, 2009

### desparate

doesnt it say P? (ie, P)?

so i think it has to be P

6. Oct 21, 2009

### tiny-tim

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.

7. Oct 21, 2009

### desparate

yes its from this book, i really dont know why we are doing this kind of physics im not studying physics or something like that

can you please draw something? i m not good at english, and i cant 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."

8. Oct 21, 2009

### tiny-tim

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.