The graph in the figure shows the variation of the electric potential V(x)(in arbitrary units)as a function of the position x (also in arbitrary units).Which of the choices below correctly describes the orientation of the x-component of the electric field along the x-axis? A)E_x is positive from x = -2 to x = 2. B)E_x is positive from x = -2 to x = 0,and negative from x = 0 to x = 2. C)E_x is negative from x = -2 to x = 0,and positive from x = 0 to x = 2. D)E_x is negative from x = -2 to x = 2.

Question

Quizplus · Accepted Answer

The electric field is related to the potential by a negative gradient, meaning $E=-dV/dx$. From the graph, we can see that the slope of the potential is positive (i.e. increasing potential) from x = -2 to x = 0, and negative (i.e. decreasing potential) from x = 0 to x = 2. Thus, the derivative of the potential changes sign at x = 0, which means that the electric field changes sign at x = 0. Therefore, E_x is positive from x = -2 to x = 0, and negative from x = 0 to x = 2.

The Graph in the Figure Shows the Variation of the Electric