Final answer:
The electric field at x = 0 m is 0 V/m, and at x = 2 m, it is -600 V/m. The electric field is found by taking the negative gradient of the electric potential V = 150x2V.
Step-by-step explanation:
The electric potential given is V = 150x2V, where x is the position along the x-axis in meters. The electric field strength, Ex, is related to the electric potential by the negative gradient of the potential. Therefore, Ex = -dV/dx. To find the electric field at a specific point, we take the derivative of the given potential with respect to x.
At x = 0 m, the derivative of V with respect to x is zero, which means that the electric field Ex at x = 0 m is also zero. At x = 2 m, the derivative of V is dv/dx = 2 × 150 × 2 m = 600 V/m, so Ex at x = 2 m is -600 V/m (negative sign indicates the direction of the field).