Find the potential between a(-7,2,1) and b(4,1,2). Given E = (-6y/x2 )i + ( 6/x) j + 5 k.

To find the potential difference between the points ( a(-7, 2, 1) ) and ( b(4, 1, 2) ) in the electric field ( mathbf{E} = left(-frac{6y}{x^2}right)mathbf{i} + left(frac{6}{x}right)mathbf{j} + 5mathbf{k} ), we can use the formula for the potential difference ( V = -int_a^b mathbf{E} cdot dmathbf{r} ).

1. Determine the Path of Integration:

We can choose a straight path in Cartesian coordinates from point A to point B.

2. Parameterize the Path:

Let ( mathbf{r}(t) = (x(t), y(t), z(t)) ).

We can parameterize it linearly:

– ( x(t) = -7 + (4 + 7)t )

– ( y(t) = 2 + (1 – 2)t )

– ( z(t) = 1 + (2 – 1)t )

where ( t ) varies from 0 to 1.

3. Differentiate the Path:

Compute ( dmathbf{r} = (dx, dy, dz) ):

– ( dx = (4 + 7)dt =