Suppose that over a certain region of space the electrical potential V is given by the following equation. V(x, y, z) = 2x2 − 3xy + xyz (a) Find the rate of change of the potential at P(2, 6, 5) in the direction of the vector v = i + j − k.

Rate of change of potential V at the point P (2,6,5) in the direction (u = i + j - k) is

(û.∇V) = 6.928

Explanation:

V(x,y,z) = (2x² − 3xy + xyz)

Rate of change of a potential, V, in a direction, u, at a point P, is given as (û.∇V)

where û = unit vector in the direction of u = (vector u)/(magnitude of u)

u = v = i + j − k

Magnitude of u = √[(1²) + (1²) + (-1)²] = √3

û = (i + j − k)/(√3)

∇V = [(∂/∂x)î + (∂/∂y)j + (∂/∂z)k](V) at point (2,6,5)

V(x,y,z) = (2x² − 3xy + xyz)

(∂V/∂x) = 4x - 3y + yz = 4(2) - 3(6) + (6)(5) = 20

(∂V/∂y) = - 3x + xz = -3(2) + (2)(5) = 4

(∂V/∂z) = xy = (2)(6) = 12

∇V = (20î + 4j + 12k)

Rate of change of potential V = (û.∇V)

û = (1/√3) (i + j − k)

(û.∇V) = (1/√3) [(i + j − k).(20î + 4j + 12k)]

(û.∇V) = (1/√3) [20 + 4 - 12) = 12/√3 = 6.928

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speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.