• Support PF! Buy your school textbooks, materials and every day products via PF Here!

It seems very hard!

  • Thread starter sma
  • Start date

sma

7
0
Consider triangle ABC,the continues charges distribution lies on the side of BC with the linear charge density such as m,prove that the direction of electric force in the point A lies on the bisector of the viewing angle BAC
 
249
0
You can prove that almost without knowing anything about electric field! We only need to know that field is determined by electric charge and that laws of physics are the same in all non-accelerated coordinate systems.

We will use this simple theory:

If we rotate the triangle (around any axis) by an angle alfa then the field will also rotate by the same angle (around the same axis). This can be proved by solving this problem in a rotated coordinate system (by alfa) where the rotated triangle seems the same as original triangle in original system (so the solution is the same). Then we transform the field vector back into original sistem and we find out it has rotated by alfa.

Proof by contadiction:
Let's suppose that the field does not lie on the bisector. If we rotate the triangle around bisector by alfa=180 degrees, then the field will also rotate, resulting in changed direction of the field. However this rotation transforms triangle back into itself! We got a different electric field from the same charge distribution! This is imposible, so the assumption that field does not lie on bisector is wrong.
 
Last edited:

sma

7
0
I think your assumpion is not right.why If we rotate the triangle around bisector by alfa=180 degrees, then rotation transforms riangle back into itself! It is true only for AB=AC!!!!!!
 
249
0
Sorry, I missread the question. Here is proof for general triangle:

Use polar coordinates, with the bisector for fi=0 axes. It is enough to show, that
sections of charged line on intervals (fi,fi+dfi) and (-fi,-fi-dfi) exactly cancel each other
out (as far as perpendicular component of E is concerned). Since sin(-fi)=sin(fi), it is enough to show that the magnitudes are the same.

Contributions of these sections are:

r^-2*dl (times a constant)

We can prove that both magnitudes of dE are the same by proving

r^-2*dl/dfi=const (independent of fi) (1)

This is easy: if delta is angle between r and BC, then

dl=r*dfi/cos(delta), r=d/cos(delta)

where d is the shortest distance between (infinitely extended) BC line and point A.

If you put dl=dfi*d/cos(delta)^2 and r=d/cos(delta) into equation (1),
you find out the expression is realy independent of fi.
 
Last edited:

sma

7
0
excuse me,I don't underestand your solution
 

malawi_glenn

Science Advisor
Homework Helper
4,782
22
Sma: You have posted this question TWICE in a very short period of time

and dont ask for solutions, and also you MUST give attempt to soultion.

Lojzek: never give full solutions

Please read and follow the rules of this forum.
 

Related Threads for: It seems very hard!

  • Posted
Replies
2
Views
5K
  • Posted
Replies
2
Views
1K
  • Posted
Replies
14
Views
3K
Replies
1
Views
10K
Replies
4
Views
9K
Replies
1
Views
8K
  • Posted
Replies
3
Views
6K
  • Posted
Replies
1
Views
1K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top