calculate the magnitude of the electric field intensity in vacuum at a distance of 20 cm from a charge of 5 * 10 raise to power - 3 column

1.1259*10^9 Newton per Columb

Explanation:

the magnitude of the electric field intensity can be calculated using the expresion below;

E=Kq/r^2

Where k= constant

q= electric charge

r=distance= 2cm= 20*10^-2m( we convert to m for unit consistency

:,K=59*10^9 Columb

If we substitute the value into above formula we have

E=(9*10^9)*(5*10^-3)/(20*10^-2)^2

=1.1259*10^9 Newton per Columb

Therefore,the magnitude of the electric field intensity in vacuum at a distance of 20 cm is 1.1259*10^9 Newton per Columb

i think its d if wrong im sorry