The electric field in a region of space is given by E_x = (3.0x) N/C, E_y = E_z = 0, where x is in m. Points A and B are on the x axis at x_A = 3.0 m and x_B = 5.0 m. Determine the potential difference V_B − V_A. A) −24 V B) +24 V C) −18 V D) +30 V E) −6.0 V

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Quizplus · Accepted Answer

The potential difference $V_B - V_A$ is calculated using the integral of the electric field over the distance from A to B. Since the electric field is given by $E = 3.0x$ N/C, and it points in the positive x-direction while the displacement from A to B is also in the positive x-direction, the potential difference is calculated as $-\int_{A}^{B} E \, dx = -\int_{3.0}^{5.0} 3.0x \, dx$. This evaluates to $-\left[\frac{3}{2}x^2\right]_{3.0}^{5.0} = -\left[\frac{3}{2}(25) - \frac{3}{2}(9)\right] = -\left[37.5 - 13.5\right] = -24$ V.

The Electric Field in a Region of Space Is Given