Question

a infinite long wire is stretched horizontally 4m above the surface of earth. it carries a charge 1 micro Columb cm inverse of its length . calculate it's electric field at a point on the earth's surface vertically below the wire.​

Answer 1

Given :

The length by which the infinite long wire is stretched above the surface of the earth = 4 m

Charge carried by this wire  = 1 micro C per cm inverse of its length

To Find :

Value of electric field at a point on the earth's surface vertically below the wire = ?

Solution :

We know that the electric field due to an infinite long wire is :

$E = \frac{2k\lambda}{r}$

Here , k is dielectric constant , $\lambda$ is charge per unit length and r is distance of point from wire .

Hence , $\lambda = 1 \frac{\mu C}{cm}$

                 =$\frac{1 \times 10^{-6}}{1 \times 10^{-2}} \frac{C}{m}$

                 =$10^{-4} \frac{C}{m}$

And , r = 4 m

Also , k =$9 \times 10^9 Nm^2 C^{-2}$

So, $E = \frac{2\times 9 \times 10 ^9 \times 10 ^{-4}}{4}$

          = $4.5 \times 10^5 \frac{N}{C}$

So, the value of electric field at a point on the earth's surface vertically below the wire is $4.5 \times 10^5 \frac{N}{C}$ .

Answer 2

Answer:

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Explanation:

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