Electrostatics Part 2

Given, length of conductor =L
Charge on conductor =Q
According to figure, OP=a=‌

√3

2

L,OQ=‌

L

2

Let ‌‌PQ=r=√OP2+OQ2
⇒‌‌PQ=√(‌

√3

2

L)2+(‌

L

2

)2=√‌

3

4

L2+‌

L

4

=L and E be the electric field at point P.
Since, E (due to finite wire) =‌

kλ

a

(sin‌φ1+sin‌φ2). . . (i)
where, k= Coulomb's constant =‌

1

4πε0

λ= linear charge density =‌

Q

L

and sin‌φ=sin‌φ2=‌

L∕2

L

=‌

1

2

Substituting the above value in Eq. (i), we get
E=‌

kλ

a

=‌

1

4πε0

‌

Q

L⋅‌ √3L 2

=‌

1

2√3πε0L2