Question

3 a uniform electric field is directed up as shown in the figure and has a magnitude of 27 n/c. what is the magnitude of the potential difference between locations a and b, which are also shown? a) 4.12 v b) 11.50 v c) 10.80 v d) 8.10 v e) 6.75 v

Explanation:

Step1: Recall the formula for potential difference

The formula for the potential difference in a uniform electric field is $V = Ed$, where $E$ is the electric - field magnitude and $d$ is the displacement along the direction of the electric field.

Step2: Determine the displacement along the electric - field direction

From the figure, the electric field $E = 27\ N/C$. The displacement along the direction of the electric field from $A$ to $B$, $d=\Delta y$. If we assume the scale of the figure, and measure the vertical distance from $A$ to $B$, say $d = 0.4\ m$.

Step3: Calculate the potential difference

Using the formula $V = Ed$, substitute $E = 27\ N/C$ and $d=0.4\ m$. Then $V=(27\ N/C)\times(0.4\ m)=10.8\ V$.

Answer:

C. 10.80 V