Potential in the x-y plane is given as volts. The electric field at the point (1,-2) will be3J V/m-5 J V/m5 J V/m-3 J V/m

To find the electric field at the point (1,-2) in the x-y plane, we need to use the concept of potential and its relationship with electric field. The electric field is the negative gradient of the potential function.

Given that the potential in the x-y plane is given as V volts, we can assume that the potential function is V(x, y), where x and y are the coordinates in the x-y plane.

To find the electric field at the point (1,-2), we need to calculate the partial derivatives of the potential function with respect to x and y, and evaluate them at the given point.

Let's denote the partial derivative with respect to x as ∂V/∂x and the partial derivative with respect to y as ∂V/∂y.

The electric field in the x-direction, Ex, is given by Ex = - (∂V/∂x), and the electric field in the y-direction, Ey, is given by Ey = - (∂V/∂y).

To find the electric field at the point (1,-2), we need to evaluate ∂V/∂x and ∂V/∂y at x = 1 and y = -2.

Once we have the values of ∂V/∂x and ∂V/∂y at the point (1,-2), we can substitute them into the equations for Ex and Ey to find the electric field at that point.

Therefore, without knowing the specific values of ∂V/∂x and ∂V/∂y, we cannot determine the exact values of the electric field at the point (1,-2).