Question

What is the magnitude of the electric field 3m away from the source if the electric potential is described by the function f(x)=2x+3, where x=2?

Answer 1

Answer:

Explanation:

Because the electric potential can be written in $V(x)=f(x)=2x+3$, then we can find:

$E=-\frac{d}{dx}V(x)=-\frac{df}{dx}=-2$ in N/C

The electric field is a constant electric field (the electric field is same for every distances from the charge) and the electric field is also comes from the negative point charge.