Question

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Find the distance from the charge. The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 and the acceleration due to gravity is 9.81 m/s 2 . Answer in units of m.

Answer 1

Answer:

The distance from the charge is 3.35 m.                

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.