Question

A ring of radius r is uniformly charged .linear charge density is .an imaginary sphere of radius r is drawn with its center on circumference

Answer 1

Answer:

Radius =r, So, 2πr= Circumference

Charge density =λ

Total charge =2πr×λ

1/3 part of the rfing is enclosed by imaginary sphere,

qin = 1/3 * 2πrλ

Фsphere = qin / ∈0

               = 2πrλ / 3∈0

                           ​

   

Explanation:

Answer 2

Answer:

$\phi = \frac{2\pi r \lambda}{3\epsilon_o}$

Explanation:

Just try to imagine the picture, what I am now going to say: We start with a ring of radius "r" which is just there with linear charge density "λ". and taking one point on the circumference as the centre of a sphere of radius same as ring ("r"), therefore this sphere passes through the centre of this ring, now you try to imagine two lines drawn from the centre of the ring to the two intersection points of the ring and sphere. Now calculate the perimeter of the ring subtended within the two lines

                                      i.e. $\frac{120}{360}\times2\pi r = \frac{2}{3} \times\pi r$

Hence, the charge inscribed inside the sphere(q) is $\frac{2\pi r}{3}\times \lambda$ (linear charge density × part of the length of the ring).

Therefore, the flux passing through the sphere is $\phi = \frac{q}{\epsilon_o} = \frac{\frac{2\pi r \lambda}{3} }{\epsilon_o}$

⇒  $\phi = \frac{2\pi r \lambda}{3\epsilon_o}$