Question

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Find an expression for electric field intensity due to a circular loop or charge..
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Answer 1

Answer:

Refer the attachment ✌️❤️

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Answer 2

Answer:

$\large\bold\red{E = \frac{1}{4\pi \epsilon_{0}} \frac{qx}{ {( {r}^{2} + {x}^{2} ) }^{ \frac{3}{2} } }}$

Explanation:

Let,

The radius of circular loop is ‘r’

And,

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 loop at a distance x from it's centre.

Therefore,

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

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

Now,

Charge on this element will be,

$\boxed{dq = ( \frac{q}{2\pi \: r} )ds}$

Clearly,

• 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 loop.

• From symmetry, this resultant field muat 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 $E_{x}$ of the field at P,

We integrate this expression over all segments of the ring.

Therefore,

We get,

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

Now,

The integral is simply the circumference of the loop which is equal to 2πr.

Therefore,

We get,

$= > \large \boxed{ \bold \pink{ E = \frac{1}{4\pi \epsilon_{0}} \frac{qx}{ {( {r}^{2} + {x}^{2} ) }^{ \frac{3}{2} } } }}$

$\purple{Note:-}$

As q is a positive charge,

Hence,

The electric field is directed away from the centre of the ring along it's axis.