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Answer: b
Explanation: Since V is given find out E.E = -Grad(V) = – Grad(10sin θ cosφ/r). From E,
we can easily compute D. D = εE = 8.854 X 10-12 X 5/2 = 22.13 units.
To find the charge density at a given point due to a potential function in a spherical coordinate system, we use the relation provided by the Poisson equation for electrostatics in vacuum or air, which is
[ nabla^2 V = -frac{rho}{epsilon_0} ]
Where ( nabla^2 ) is the Laplace operator (in this case, in spherical coordinates), (V) is the potential, ( rho ) is the charge density, and ( epsilon_0 ) is the permittivity of free space ((8.854 times 10^{-12} , C^2/N cdot m^2 )).
Given, ( V = frac{10sin(theta)cos(phi)}{r} )
First, let’s apply the Laplacian in spherical coordinates to (V), remembering that (r), (theta), and (phi) are the radius, polar angle, and azimuthal angle, respectively. The Laplacian of a scalar field (V(r, theta, phi)) in spherical coordinates is given by:
[ nabla^2 V = frac{1}{r^2} frac{partial}{partial r} left( r^2 frac{partial V}{partial r} right) + frac{1}{r^2 sin