Calculate the magnitude of the electric field intensity in a vacuum at a distance of 25cm from a charge of 5 × 10-3C. Take (\(\frac{1}{4πε0} = 9\times10^9Nm^2C^{-2}\))

Rhauph5

5 years ago

Electric field intensity = kq/d^2
where K- constant 1/4πε0 = 9*10^9
q- charge in Columb
D- distance in meters
E={ (9*10^9) * (5 × 10^-3)}/ (0.25)^2
= 7.2 × 10^8NC^-1

nonoimeh

5 years ago

kqq/d^2
k is de same thing as 1/4π£°
K=9*10^9
Changing cm to m
25/100=0.25m
So 9*10^9 * 5*10^-3/0.25^2
=7.2*10^8NC^-1

Sensei

2 months ago

type fvckin shi