Find the potential between two points p(1,-1,0) and q(2,1,3) with E = 40xy i + 20×2 j + 2 k

To find the potential difference (V) between two points (P(1, -1, 0)) and (Q(2, 1, 3)) in an electric field described by the vector field (mathbf{E} = 40xy mathbf{i} + 20x^2 mathbf{j} + 2 mathbf{k}), we use the formula for the electric potential difference:

[

V = -int_P^Q mathbf{E} cdot dmathbf{l}

]

where (dmathbf{l} = dxmathbf{i} + dymathbf{j} + dzmathbf{k}) is an infinitesimal displacement vector along the path from (P) to (Q). For the sake of simplicity, let’s take the path of integration to be straight from (P) to (Q).

The electric field vector is given by:

[

mathbf{E} = 40xy mathbf{i} + 20x^2 mathbf{j} + 2 mathbf{k}

]

Given the points:

– (P(1, -1, 0))

– (Q(2, 1, 3))

We find the change in coordinates from (P) to (Q):

– (Delta x = 2 – 1 = 1)

– (