Given,
E = 40xy i + 20x^2 j + 2k
p(1,-1,0) and q(2,1,3)

To find the potential between the two points, we integrate the given E field with respect to distance along the path between the two points.

Potential difference between two points A and B is given by:
VAB = -∫E.dr (From point A to point B)

where E is the electric field between points A and B and dr is the infinitesimal displacement vector along the path from A to B.

Let's first find the electric field between the two points.

Electric field E = -∇V

where V is the potential and ∇ is the del operator.

Since the electric field is the gradient of the potential, we can find the potential by integrating the electric field along the path between the two points.

∫E.dr = -∫∇V.dr = -(V(q) - V(p))

where V(q) and V(p) are the potentials at points q and p respectively.

∫E.dr = -(V(q) - V(p)) = V(p) - V(q)

Let's find the potential at point p and q.

At point p,
x = 1, y = -1, z = 0
Potential at point p = V(p) = ∫E.dr = ∫(40xy i + 20x^2 j + 2k).dr

Let's assume the path from p to q is along the x-axis.

dr = dx i + 0 j + 0 k
x varies from 1 to 2

∫E.dr = ∫(40xy i + 20x^2 j + 2k).(dx i + 0 j + 0 k) = ∫(40xy)dx = ∫(40*1*(-1))dx = -40∫dx = -40(x)|^2_1 = -40(2 - 1) = -40

Potential at point p = V(p) = ∫E.dr = -(-40) = 40

At point q,
x = 2, y = 1, z = 3
Potential at point q = V(q) = ∫E.dr = ∫(40xy i + 20x^2 j + 2k).dr

Let's assume the path from q to p is along the x-axis.

dr = dx i + 0 j + 0 k
x varies from 2 to 1

∫E.dr = ∫(40xy i + 20x^2 j + 2k).(dx i + 0 j + 0 k) = ∫(40xy)dx = ∫(40*2*1)dx = 80∫dx = 80(x)|^1_2 = 80(-1) = -80

Potential at point q = V(q) = ∫E.dr = -(-80) = 80

Potential difference between point p and q = V(p) - V(q) = 40 - 80 = -40

Therefore, the potential between the two points is 40 volts.

But the given options do not match the answer. So let's check the mistake.

The correct answer is obtained by taking the