Find the electric field at the centre of a uniformly charged semi-circular arc of radius 'a' and linear charge density 'lambda'.
Answers
Answered by
9
Answer:
it is very big so can't explain here sorry
Explanation:
i know everything about this.
Answered by
0
Answer:
Correct option is
A
2πε
0
R
λ
Consider an element having charge dq at angle θ of angular width dθ.
Length of element=Rdθ
dq=λRdθ
Field due to this element=dE=
4πϵ
o
R
2
dq
⟹dE=
4πϵ
o
R
2
λRdθ
=
4πϵ
o
R
λdθ
Now, from symmetry we can see that the horizontal component of field of right side get cancelled by that of left side.
And, the vertical component of force add up.
E=∫dE
y
=∫dEsinθ
⟹E=∫
0
π
4πϵ
o
R
λ
sinθdθ
⟹E=
4πϵ
o
R
λ
[−cosθ]
0
π
⟹E=
4πϵ
o
R
2λ
⟹E=
2πϵ
o
R
λ
Answer-(A)
solution
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