Consider a uniform horizontal $\vec { E }$ field of 50 N/C directed toward the east.If the electric potential at a given point in the field is 80 V,what is the potential at a point 1.0 m directly south of that point?

Question

Quizplus · Accepted Answer

The electric field and electric potential are related by the equation:
$\vec{E}=-\frac{d\vec{V}}{dx}$
where $\vec{E}$ is the electric field, $\vec{V}$ is the electric potential, and x is the distance in the direction of the electric field.

Since the electric field is directed toward the east and the potential is 80 V, we know that the change in potential in one meter in the direction of the field is:
$-\frac {dV}{dx} = E = 50 N/C$
$\Delta V = -E\Delta x = -(50 N/C)(1.0 m) = -50 V$

Therefore, the potential at a point 1.0 m directly south of the original point is:
$V = 80 V + \Delta V = 80 V - 50 V = 30 V$

The correct answer is (B) 30 V.

Consider a Uniform Horizontal E⃗\vec { E }E Field of 50 N/C Directed Toward the East

Consider a Uniform Horizontal $\vec { E }$ Field of 50 N/C Directed Toward the East