Derive an expression for the Electric field intensity at the centre of an arc of radius R and having a charge Q uniformly distributed on it where , the arc subtends an angle 
at the centre.
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Heyy Here is your answer !!
Answer is long as I explained at every part !
Sorry while writing your earlier answer ,I couldnt resist myself to left this unanswered question:
See pic 2 to see how figure looks like .
For figure Clarification :
My drawing is very poor !!
1) I had divided the arc into two symmetrical parts that make angle α/2 with respect to horizontal .
Pre-requiste to understand my answer :
1) Electric field intensity due to a point charge dq is
2) Basic Integration like sinx and cos x .
In my pic, R (r) is radius of curve
Cartesian 2 D co -ordinate system :
x is along right side and y is along upward direction:
Steps :
1) In symmetrical figure, consider a charge 'dq ' at an angle ( theta ) above a horizontal (here theta is variable ) .
=>Electric field due to charge 'dq ' which is above horizontal is
Replace theta by angle 'a'.
dE(1) = dE cos (a) i -dE sin (a) j
where i and j are along positive x,y direction.
Similarly,
Consider a point charge. (dq) below the horizontal at angle theta or (a).
Now,
Electric Field due to this charge
is
dE(2) = dE cos (a)i +dE sin (a) j
Since ,
on adding
dE (1) + dE(2) 'y' component gets cancelled out :
Since, this sum is taken at some general angle
(a), therefore we can say that net Electric field of the arc (whole) will be along positive x direction:
Or
You can also imagine it as for every charge dq above the horizontal, there exits a charge dq which cancels its effect along y axis .
From, here we can say that Electric Field Intensity of circular arc will be along direction x
where x direction is taken as the line joining the mid -pointvof the arc to the center .
Step (1) is just for your visualization of question and direction of net Electric field.
2) Now, coming back to the Question :
Consider a charge dq above horizontal at angle (a) or theta as mentioned in pic .
Electric field due to this charge be dE.
Electric field due to this charge along x -axis is
dEcos (a) and forget about y direction :
As we know net Electric field will be summation of Electric field along x direction of all dq charges on the arc as summation of Electric field along y direction gets cancelled out:
linear charge density (In pic it is lambda)
= Q/ rα
dq = Lambda r da ( see on right side of pic)
a is angle as theta .
Now finally,
dE (x) = dE cos (a) i
dE (x) = kdq cos (a) /r^2
For Calculation or Integration see pic :
Assuming α angle is in radian :
Final Result : E = 2 kQ sin (α/2) / ( r ^2 α)
Hope you understand my answer and it may helps you !
If answered it , as by mistake I checked your Questions Section where this Question is unanswered
:
So, I couldnt resist myself to left it as it is:
Sorry , for my bad writing and drawin
If you dont understand my answer or there is an error , Report it Freely !
Answer is long as I explained at every part !
Sorry while writing your earlier answer ,I couldnt resist myself to left this unanswered question:
See pic 2 to see how figure looks like .
For figure Clarification :
My drawing is very poor !!
1) I had divided the arc into two symmetrical parts that make angle α/2 with respect to horizontal .
Pre-requiste to understand my answer :
1) Electric field intensity due to a point charge dq is
2) Basic Integration like sinx and cos x .
In my pic, R (r) is radius of curve
Cartesian 2 D co -ordinate system :
x is along right side and y is along upward direction:
Steps :
1) In symmetrical figure, consider a charge 'dq ' at an angle ( theta ) above a horizontal (here theta is variable ) .
=>Electric field due to charge 'dq ' which is above horizontal is
Replace theta by angle 'a'.
dE(1) = dE cos (a) i -dE sin (a) j
where i and j are along positive x,y direction.
Similarly,
Consider a point charge. (dq) below the horizontal at angle theta or (a).
Now,
Electric Field due to this charge
is
dE(2) = dE cos (a)i +dE sin (a) j
Since ,
on adding
dE (1) + dE(2) 'y' component gets cancelled out :
Since, this sum is taken at some general angle
(a), therefore we can say that net Electric field of the arc (whole) will be along positive x direction:
Or
You can also imagine it as for every charge dq above the horizontal, there exits a charge dq which cancels its effect along y axis .
From, here we can say that Electric Field Intensity of circular arc will be along direction x
where x direction is taken as the line joining the mid -pointvof the arc to the center .
Step (1) is just for your visualization of question and direction of net Electric field.
2) Now, coming back to the Question :
Consider a charge dq above horizontal at angle (a) or theta as mentioned in pic .
Electric field due to this charge be dE.
Electric field due to this charge along x -axis is
dEcos (a) and forget about y direction :
As we know net Electric field will be summation of Electric field along x direction of all dq charges on the arc as summation of Electric field along y direction gets cancelled out:
linear charge density (In pic it is lambda)
= Q/ rα
dq = Lambda r da ( see on right side of pic)
a is angle as theta .
Now finally,
dE (x) = dE cos (a) i
dE (x) = kdq cos (a) /r^2
For Calculation or Integration see pic :
Assuming α angle is in radian :
Final Result : E = 2 kQ sin (α/2) / ( r ^2 α)
Hope you understand my answer and it may helps you !
If answered it , as by mistake I checked your Questions Section where this Question is unanswered
:
So, I couldnt resist myself to left it as it is:
Sorry , for my bad writing and drawin
If you dont understand my answer or there is an error , Report it Freely !
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