The electric field in the region defined by the y-z plane and the positive x axis is given by E = ax,where a is a constant.(There is no field for negative values of x. )As x increases in magnitude,relative to V = 0 at the origin,the electric potential in the region defined above is A)a decreasing function proportional to -|x²|. B)a decreasing function proportional to -|x|. C)constant. D)an increasing function proportional to +|x|. E)an increasing function proportional to +|x²|.

Question

Quizplus · Accepted Answer

The electric potential is given by the negative integral of the electric field. In this case, the electric field is in the x-direction only, so the potential is a function of x only. The potential is zero at the origin, so the potential at a point x is given by:$$V(x) = -\int_0^x E dx = -\int_0^x a dx = -ax^2/2$$Therefore, the potential is a decreasing function proportional to |x^2|.

The Electric Field in the Region Defined by the Y-Z