In a region where the electric field is uniform and points in the +x direction,the electric potential is -2000 V at x = 8 m and is +400 V at x = 2 m.What is the magnitude of the $\vec { E }$ field?

Question

Quizplus · Accepted Answer

The electric field is related to the change in electric potential with distance according to:

$\vec{E}=-\frac{\Delta V}{\Delta x}$

Given that the electric potential decreases by 2000 V over a displacement of 8 m in the +x direction, and increases by 400 V over a displacement of 2 m in the same direction, we have:

$\Delta V = 400\mathrm{V}-(-2000\mathrm{V})=2400\mathrm{V}$
$\Delta x = 2\mathrm{m}-8\mathrm{m}=-6\mathrm{m}$

Substituting into the formula for the electric field, we get:

$\vec{E}=-\frac{\Delta V}{\Delta x}=-\frac{2400\mathrm{V}}{-6\mathrm{m}}=400\mathrm{V/m}$

Therefore, the magnitude of the electric field is 400 V/m, which corresponds to choice C).

In a Region Where the Electric Field Is Uniform and Points