Question

Electric field intensity of a point at a distance of 2000 Aº from an a particle is
NC-1

Answer 1

Given:

Point of 2000 A° from a charge ?

To find:

Electric field intensity ?

Calculation:

Let the charge have magnitude of q.

Now, field intensity will be :

$\rm E = \dfrac{kq}{ {r}^{2} }$

$\rm \implies E = \dfrac{9 \times {10}^{8} \times q }{ {(2000 \times {10}^{ - 10}) }^{2} }$

$\rm \implies E = \dfrac{9 \times {10}^{8} \times q }{ {(2 \times {10}^{ - 7}) }^{2} }$

$\rm \implies E = \dfrac{9 \times {10}^{8} \times q }{4 \times {10}^{ - 14} }$

$\rm \implies E = 2.25 \times {10}^{22} q \: \: \: N/C$

So, field intensity is 2.25 × 10²²q N/C.