Answer:
Sure! Let's address each part of your question:
i) To check if the electric field F is conservative, we need to verify if it satisfies the condition of being a curl-free field. In other words, we check if the curl of F is zero. Let's calculate the curl of F:
∇ × F = (∂Fy/∂x - ∂Fx/∂y) i + (∂Fx/∂x - ∂Fy/∂y) j
For F = xy^2 i + x⋅y j, we have:
∂Fy/∂x = 0
∂Fx/∂y = 0
∂Fx/∂x = 0
∂Fy/∂y = 0
Hence, the curl of F is zero, which implies that the electric field F is conservative.
To find a potential function for F, we integrate the components of F with respect to the respective variables:
∫Fx dx = ∫xy^2 dx = (1/2)xy^2 + g(y)
∫Fy dy = ∫x⋅y dy = (1/2)x⋅y^2 + h(x)
Since F is conservative, the potential function φ(x, y) is given by:
φ(x, y) = ∫Fx dx + ∫Fy dy = (1/2)xy^2 + (1/2)x⋅y^2 + C
where C is the constant of integration.
ii) To evaluate the work done by the electric field Ē in moving a particle from (0,0) to (1,1) along any two different paths, we can use the work-energy theorem. The work done (W) is given by:
W = ΔU = φ(1, 1) - φ(0, 0)
Substituting the values, we get:
W = [(1/2)(1)(1)^2 + (1/2)(1)(1)^2 + C] - [(1/2)(0)(0)^2 + (1/2)(0)(0)^2 + C]
= 1/2 + C - 0
= 1/2 + C
Therefore, the work done by the electric field Ē in moving a particle from (0,0) to (1,1) along any two different paths is 1/2 + C, where C is the constant of integration.
iii) The potential difference between two points in an electric field is the difference in their potential function values. Here, we need to find the potential at (1,0) and (2,1).
Using the potential function φ(x, y) = (1/2)xy^2 + (1/2)x⋅y^2 + C, we can evaluate the potential at (1,0) and (2,1):
Potential at (1,0):
φ(1, 0) = (1/2)(1)(0)^2 + (1/2)(1)(0)^2 + C
= 0 + 0 + C
= C
Potential at (2,1):
φ(2, 1) = (1/2)(2)(1)^2 + (1/2)(2)(1)^2 + C
= 1 + 1 + C
= 2 + C
Therefore, the potential difference between the points (1,0) and (2,1) is:
Potential at (2,1) - Potential at (1,0) = (2 + C) - C = 2.
iv) To explain the results, we can conclude the following:
- The electric field F is conservative since its curl is zero.
- The potential function φ(x,y) for F is given by (1/2)xy^2 + (1/2)x⋅y^2 + C, where C is a constant of integration.
- The work done by the electric field in moving a particle from (0,0) to (1,1) along any two different paths is determined by the potential difference, which is 1/2 + C.
- The potential difference between the points (1,0) and (2,1) is 2.
These results indicate that the electric field under consideration possesses conservative properties, allowing for the existence of a potential function. The specific values of the potential and potential difference rely on the constant of integration C and the coordinates of the points involved.