If the electric field is in the positive x direction and has a magnitude given by E = Cx², where C is a constant, then the electric potential is given by V =

Question

Quizplus · Accepted Answer

The electric potential is given by V = -∫E dx, where the integral is taken along the path of interest. Since the electric field is in the positive x direction, the path of interest is in the x direction. Thus, V = -∫C x^(1+(1/2)+1)(2+(1/2)+1)/(1+(1/2)) dx from 0 to x. Simplifying the integral and evaluating the limits, we get V = -Cx^(5/2)/(5/2) from 0 to x, which simplifies to V = -2Cx^(5/2)/5. To find the potential difference between two points, we subtract the potential at the lower point from the potential at the higher point. Let the two points be x = 0 and x = L. Then the potential difference is ΔV = (-2C(L)^(5/2)/5) - (-2C(0)^(5/2)/5), which simplifies to ΔV = -2CL^(5/2)/5. Combining with the expression for the electric field, we find that ΔV = -Cx^(3/2)^/(3), which simplifies to ΔV = -Cx^(1+1/2)^/(3), which is the same as the expression given for choice D.

If the Electric Field Is in the Positive X Direction