Question

can anyone find electric feild on ring on its àxial line​

Answer 1

Answer:

$\large\boxed{\sf{E = \dfrac{1}{4\pi{\epsilon}_{0}}\dfrac{qx}{{({r}^{2}+{x}^{2})}^{\frac{3}{2}}}}}$

Explanation:

Let, the radius of circular ring is ‘r’

Also, Let's assume that,

The total charge uniformly distributed is ‘q’

Now, we have to find the electric field at a point P that lies on the axis of the ring at a distance x from it's centre.

Therefore, we have to,

Consider a differential element of the ring of length ‘ds’.

$\red{Note:-}$ Refer to the attachment for figure.

Now, we have,

Charge on this element will be,

$\large{dq = ( \dfrac{q}{2\pi r}) ds}$

Clearly, we have,

This element sets up a differential electric field dE at point P.

The resultant field, E is found by integrating the effects of all the elements that makes up the ring.

From symmetry, this resultant field must lie along the right axis.

Thus only one component of dE parallel to this axis contributes to the final result.

Now, to find the total x- component of of the field at P,

We integrate this expression over all segments of the ring.

Therefore, we will get,

$=>E_{x} = \displaystyle \int dE \cos \alpha$

$=>E_{x} = \displaystyle \int \dfrac{k \: dq}{( {r}^{2} + {x}^{2} )} \dfrac{x}{ \sqrt{ {r}^{2} + {x}^{2} } } \\ \\ = > E_{x} = \dfrac{kqx}{2\pi r {( {r}^{2} + {x}^{2} )}^{ \frac{3}{2} } } \displaystyle \int ds$

But, the integral is simply the circumference of the ring which is equal to 2πr.

Hence, We will get,

The electric field for ring on it's axial line is :

$\large\boxed{\sf{E = \dfrac{1}{4\pi{\epsilon}_{0}}\dfrac{qx}{{({r}^{2}+{x}^{2})}^{\frac{3}{2}}}}}$