Question

Find the intensity of electric field at a distance
of 0.2m from the charge of 5MC.​

Answer 1

Answer:

Explanation:

E = kQ/$r^{2}$

=  (9 x $10^{9}$) (5 x $10^{6}$) / (4 x $10^{-2}$)

1.125 x $10^{-18}$ N/C

Answer 2

Given:

The charge Q = 5 mega coloumb

                        = 5 × 10 ⁶ C

To Find:

the intensity of the electric field at a distance of 0.2m from the charge of

 5 × 10 ⁶ C.

Solution:

As we know,

The electri intensity at distance r from charge Q is given by.

E = kQ/ r²

Where k = 8.99 x 10⁹ N m²/C²

E = ( 9 x 10⁹ N m²/C² × 5 × 10 ⁶ C )/ ( 0.2 )²

E = 1125 x 10¹⁵

E = 1.125  x 10¹⁸ N/C

Hence, the intensity of electric field is 1.125  x 10¹⁸ N/C.