Question

What is the magnitude of the electric field in a region where the potential is given by the expression

V = ax² + b where a = −150 V/m² and b = 175 V?

Answer 1

The magnitude of the electric field in a region with a given potential can be calculated using the equation E = -dV/dx. In this case, the potential is given by V = ax^2 + b. By differentiating the potential with respect to position, the electric field is found to be 300x V/m.

Step-by-step explanation:

The electric field, E, is related to the potential, V, by the equation:

E = -dV/dx

where dV is the change in potential along the x-direction and dx is the change in position along the x-direction.

In this case, the potential is given by: V = ax^2 + b

So, the electric field can be calculated as:

E = -dV/dx = -d/dx(ax^2 + b)

Since b is a constant, the derivative of b with respect to x is 0. Therefore, the electric field becomes:

E = -d/dx(ax^2) = -2ax

Substituting the given values of a and b:

E = -2(-150)x = 300x

Thus, the magnitude of the electric field is 300x V/m.