Question

The electric field intensity at a point 2 m from an isolated point charge is 500 N/C. The electric potential at the point is?

Answer 1

Answer:

I think sooooooooooooo it's 1000

Answer 2

Given,

electric field intensity = 500 N/C

r = 2 m

To find,

Electric potential at a point

Solution,

Electric field intensity is given as

$E = \frac{1}{4\pi e}$×$\frac{q}{r^{2} }$

E = 500 N/C

Electric potential is given as:

V = 1/4πε × q/r

we can also give electric potential as

V = (1/4πε) × (q/r²) × r

V = E × r

V = 500 × 2

V = 1000

Therefore electric potential at isolated point is 1000 V.