Question

Ring of radius r having linear charge density lambda moves towards a solid imaginary sphere of radius r by 2 so that the centre of the ring passes through the centre of the sphere the axis of the ring is perpendicular to the line joining the centre of the ring and the sphere the maximum flux through the sphere in this process is

Answer 1

Answer:  The correct answer is λ $\frac{r\pi }{3e_0}$ .

Explanation:

Given data in the question,

Radius of Ring = r , Radius of sphere = $\frac{r}{2}$

To find maximum charge in the sphere, we have the formula,

q = 2 . λ . Radius

∴ q = 2 ( $\frac{r}{2}$ ) . λ

∴ q = rλ  

Maximum flux through the sphere in the process is calculated as

F = Ф =  $\frac{q}{e_0}$    .................equation 1

When the centre of ring will pass through the centre of the sphere,

the angle subtended will be $\frac{\pi }{3}$ at the centre of the ring.

∴ charge on this arc = q = $\frac{r\pi }{3}$ λ   ...............equation 2

putting the value of q in equation 1, we get

Ф = = $\frac{\frac{r\pi }{3} }{e_0}$ λ

Ф = λ $\frac{r\pi }{3e_0}$

∴ Maximum flux through the sphere in the process is λ $\frac{r\pi }{3e_0}$ .

Answer 2

The value of maximum flux through the sphere is Ф =  λ x r π / 3εo

Explanation:

Given data:

• Radius of Ring = r
• Radius of sphere = r / 2

To find maximum charge in the sphere, we have the formula,

q = 2 . λ . Radius  

q = 2 (r / 2) . λ

q = rλ  

Maximum flux through the sphere in the process is calculated as

F = Ф = q / εo  -------- (1)

Charge on arc = q = r π / 3 λ     --------(2)

Now put the value of q

Ф = r π / 3 / εo x λ

Ф =  λ x r π / 3εo

Hence the value of maximum flux through the sphere is Ф =  λ x r π / 3εo