Question

Find the electric field at the centre of a uniformly charged semi-circular arc of radius 'a' and linear charge density 'lambda'.​

Answer 1

Answer:

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 2

Answer:

join for =xjn-pfkv-nxt