jangyasinniTeacher
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a
Explanation: The Poisson equation is given by Del
2
(V) = -ρ/ε. To find ρ, put ε = 8.854 x
10
-12 in air and Laplacian of the function is 2 + 2 + 2 = 6. Ρ = 6 x 10
-9
/36π = 1/6π units.
To calculate the charge density when given a potential function (V = x^2 + y^2 + z^2) in air, we need to use the relation from electrostatics that links the electric potential (V) to charge density (rho). In vacuum or air (for the sake of this problem, we assume air is a vacuum for simplicity), the relation is given by Poisson’s equation,
[
nabla^2 V = -frac{rho}{epsilon_0}
]
where (nabla^2) is the Laplacian operator, (V) is the electric potential, (rho) is the charge density, and (epsilon_0) is the vacuum permittivity (electric constant), with a value approximately equal to (8.854 times 10^{-12} , text{C}^2/text{N}cdottext{m}^2).
Given (V = x^2 + y^2 + z^2), we can calculate the Laplacian (nabla^2 V):
[
nabla^2 V = frac{partial^2 V}{partial x^2} + frac{partial^2 V}{partial y^2} + frac{partial^2 V}{partial z^2}
]
Substituting (V) into the formula gives:
[