What is the potential difference between 10sinθcosφ/r 2 at A(1,30,20) and B(4,90,60)?

To solve this question and find the potential difference between the two points, we’ll interpret the potential function as (V(x,y,z) = frac{10sinthetacosphi}{r^2}), where (r) is the distance from the point of reference, usually the origin, (x) is represented as (rsinthetacosphi), (y) as (rsinthetasinphi), and (z) as (rcostheta). These are the spherical coordinates transformation equations. Given points A and B are provided in spherical coordinates as (A(r,theta,phi) = A(1, 30°, 20°)) and (B(r,theta,phi) = B(4, 90°, 60°)).

First, let’s evaluate the potential at point A ((V_A)):

– (r_A = 1)

– (theta_A = 30°)

– (phi_A = 20°)

[V_A = frac{10sin(30°)cos(20°)}{1^2}]

To convert the angles into radians which is often the required format for mathematical functions in calculators and programming languages:

– (30° = frac{pi}{6}) radians and (20° = frac{pi}{9}) radians approximately.

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