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Chapter 14: Problem 6

What is the divergence of an inverse square vector field?

Short Answer

Expert verified
Answer: The divergence of an inverse square vector field is zero.

Step by step solution

01

1. Define the inverse square vector field

An inverse square vector field is given by the expression: \(\vec{F}(x, y, z) = \frac{A}{r^3}\vec{r}\), where \(A\) is a constant, \(r = \sqrt{x^2 + y^2 + z^2}\) is the distance from the origin, and \(\vec{r} = (x, y, z)\) is the position vector.
02

2. Write down the general formula for the divergence of a vector field

The divergence of a vector field \(\vec{F}\), denoted as \(\nabla \cdot \vec{F}\), is given by the sum of the partial derivatives of each component of the field: \(\nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}\).
03

3. Write down the components of the inverse square vector field

The components of the inverse square vector field are given by: \(F_x = \frac{A}{r^3}x\), \(F_y = \frac{A}{r^3}y\), \(F_z = \frac{A}{r^3}z\).
04

4. Compute the partial derivatives of the components of the vector field

We compute the partial derivatives of the components of the vector field with respect to \(x\), \(y\), and \(z\): \(\frac{\partial F_x}{\partial x} = \frac{A(3x^2 - r^2)}{r^5}\), \(\frac{\partial F_y}{\partial y} = \frac{A(3y^2 - r^2)}{r^5}\), \(\frac{\partial F_z}{\partial z} = \frac{A(3z^2 - r^2)}{r^5}\).
05

5. Compute the divergence of the inverse square vector field

We substitute the computed partial derivatives into the expression for the divergence and sum them: \(\nabla \cdot \vec{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z} = \frac{A(3x^2 - r^2)}{r^5} + \frac{A(3y^2 - r^2)}{r^5} + \frac{A(3z^2 - r^2)}{r^5}\). Simplifying the expression, we get: \(\nabla \cdot \vec{F} = \frac{A(3x^2 + 3y^2 + 3z^2 - 3r^2)}{r^5} = \frac{A(3r^2 - 3r^2)}{r^5} = 0\). Thus, the divergence of an inverse square vector field is zero.

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