Question

Electric field intensity is 400Vm^-1 at a distance 2m from point charge . It will be 100Vm^-1 at a distance?

Answer 1

Answer  electric potential=electric field*distance

                                             =400*2=800 v

             distance when electric field intensity is 100=800/100=8 m

Explanation:if it helps mark as brainliest

Answer 2

Answer:

4 m

Explanation:

we know that E=$\frac{Kq}{r^{2} }$

∴Kq=400×2²

acc to question

100=$\frac{400*2^{2} }{r^{2} }$

⇒r=4