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 (vec{E} = 40xy vec{i} + 20x^2 vec{j} + 2 vec{k},) we use the formula:

[V = – int_{P}^{Q} vec{E} cdot dvec{r},]

where (dvec{r}) represents an infinitesimal displacement vector in the field, and (cdot) denotes the dot product.

First, let’s parametrize the path from (P) to (Q). A straightforward path is a line that can be described by parametric equations. Given points (P(x_1, y_1, z_1) = (1, -1, 0)) and (Q(x_2, y_2, z_2) = (2, 1, 3)), we can find parameters for the line connecting these points. The parametric line can be represented as (vec{r}(t) = vec{r}_0 + tvec{d},) where (vec{r}_0) is the initial point vector, (vec{d}) is the direction vector from (P) to (Q),