Question

The potential due to a point charge at a distance of
0.3 m is 10^ 3.
What will be the electric field intensity
due to the same charge at a distance of 0.4 m?
Ans. 1875 V-m
-1​

Answer 1

Answer:

1875V-m^-1

Step-by-step explanation:

We have V=Kq/r = 1000 volt. So Kq = 1000 r = 1000 x 0.3 = 300 units.

E = Kq/(R^2) = 300 x (100/16) = 1875 N/C.

Answer 2

Answer:

1000*0.3=300

Kq=300 then kq/r^2 is

E=300/(0.4)^2

1875v-m. answer