TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Given, Radius of loop, r=10cm=10×10−2m
Total charge on loop, q=10−5C
Small arc length, x=3.14×10−6m
Let, E be the electric field.
As we know that,
E=‌

kq′

r2

. . . (i)
where, k= Coulomb's constant
and q′= charge on arc length x.
∴‌‌q′‌‌=‌

q

2πr

x
‌‌=‌

10−5

2π×10×10−2

×314×10−6=‌

1

2

×10−10C
Substituting in Eq. (i), we get
E‌‌=‌

9×109× 1 2 ×10−10

(10×10−2)2

‌‌=4.5×109−10+2=45N∕C