Given: Ē = 2y î 2x j

To find: V(x,y,z)

Solution:

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

E = -∇V

where ∇ is the gradient operator.

To find V, we need to integrate E with respect to position:

V(x,y,z) = -∫E·ds

where the integral is taken along any path from an arbitrary reference point to the point (x,y,z).

Steps:

1. Integrate E along the x-axis:

V(x,y,z) = -∫E·ds = -∫(2y î)·dx = -2xy + C1(y,z)

where C1(y,z) is an arbitrary constant of integration.

2. Integrate E along the y-axis:

V(x,y,z) = -∫E·ds = -∫(2x j)·dy = -2xy + C2(x,z)

where C2(x,z) is another arbitrary constant of integration.

3. Combine the two integrals and add an additional constant of integration C3(z):

V(x,y,z) = -2xy + C1(y,z) - 2xy + C2(x,z) + C3(z)

4. Simplify the expression by combining the two constants of integration:

V(x,y,z) = -4xy + C(y,z,x)

where C(y,z,x) is an arbitrary constant of integration.

Final Answer:

Therefore, the potential difference V(x,y,z) is given by:

V(x,y,z) = -4xy + C(y,z,x)

where C(y,z,x) is an arbitrary constant of integration.