AP EAPCET 06-Jul-2022 Shift 1 Paper

The given situation is shown below.On straight wire λ= charge per unit length =Q∕L
small charge on length element dl is
dq=λ×dl .....(i)
Electric field at centre due to a point charge
=K×‌

dq

r2

When each electric line is broken into 2 components, horizontal component of electric field balance each other, only vertical components remain.
So, net electric field along vertical
dE‌vertical ‌‌=2dE‌cos‌θ=2‌

Kdq

r2

‌cos‌θ
‌=‌

2K‌cos‌θ

r2

×λdl( But dθ=‌

dl

r

⇒dl=rdθ)
Integrating the electric field,
E‌=

‌ π 2

∫

0

‌

2Kλ‌cos‌θ‌d‌θ

r

but length of semi-circular path is
L‌=πr
⇒‌‌r‌=‌

L

π

∴‌‌E‌=‌

2KQ

L×‌ L π

=‌

2KπQ

L2

∴E‌=‌

2

4πε0

×‌

πQ

L2

=‌

Q

2ε0L2

.