Question

Find electric field at point o due to segment of a ring while linear charge density is 8c/cm o 60 degree and distance is 4 cm

Answer 1

Answer with explanation

Linear Charge Density

$=\lambda=\frac{8\times Coulomb}{cm}$

Angle between ,length vector and line from center of the ring to end of segment which is in the shape of wire =60°

Let,a be the radius of the ring.

And, Distance from center of the ring to end of wire(X) = 4 cm

R=Distance between center of ring to end of wire which is in the form of segment

cos 60°

        $=\frac{X}{R}\\\\ \frac{1}{2}=\frac{4}{R}\\\\ R=4 \times 2\\\\R=8$

R=8 cm

R²=a²+X²

8²=4²+a²

64-16=a²

a²=48

a=√48

a=4√3 cm

$\lambda=\frac{Q}{2\times \pi \times a}\\\\ 8=\frac{Q}{2\times 3.14 \times 4\sqrt{3}}\\\\ Q=6.28 \times 4 \times 1.732\times 8\\\\ Q=348.06272$

Total charge inside the ring =348.06 Coulomb(approx)

Electric field produced at a distance of 4 cm ,by the ring

  $=\frac{Q\times X}{4\pi\times \epsilon\times R^3}=\frac{348.06 \times 4\times 9\times 10^9}{8^3}{\text{because}}4\times\pi\epsilon=\frac{1}{9\times 10^9\\\\=24.47296875\times 10^9=2.45\times 10^10$

So, total electric field , produced by the ring at a distance of 4 cm ,while linear charge density is 8$\frac{Coulomb}{cm}\\\\ =2.45\times 10^10\frac{Newton}{Coulomb}$

Answer 2

Answer:

FORMULA FOR ELECTRIC FIELD DUE TO RING=

E=Lamba/2 Pi epsilon R Sin theta/2.

Sorry for CROP IMAGE.

ALL THE BEST FOR JEE.