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

Question

Quizplus · Accepted Answer

The potential difference ($\Delta V$) in an electric field ($\vec{E}$) is given by $\Delta V = -\vec{E} \cdot \vec{d}$, where $\vec{d}$ is the displacement vector. Here, $\vec{E} = 50\, \text{N/C}$ east, and $\vec{d} = 1.0\, \text{m}$ west. Since the displacement is opposite to the field direction, $\Delta V = 50\, \text{N/C} \times 1.0\, \text{m} = 50\, \text{V}$. The potential increases when moving against the field, so the potential 1.0 m west is $80\, \text{V} + 50\, \text{V} = 130\, \text{V}$.

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