Question

Find the divergence of the following vector field: x y b2 r b 2 r b2 r and r Vx y z where bis a constant

please explain the math - my calc is extreamely rusty

Answer 1

The divergence of a vector is $\nabla . \vec{A} = \frac{\partial A_x}{\partial x} + \frac{\partial A_y}{\partial y} + \frac{\partial A_z}{\partial z}$ , so we need to take the derivative of each term of the field respect to the coordinate. So lets start with this: $\frac{\partial E_x}{\partial x} = \frac{\partial \frac{x}{b^{2} + r^{2}}}{\partial x} = \frac{\partial }{\partial x} \bigg(\frac{x}{b^{2} + r^{2}}\bigg)$ , this is the derivative of a division due to the fact that "r" is a function of "x". Therefore we use $\frac{\partial }{\partial x} \bigg(\frac{f(x)}{g(x)}\bigg) = \frac{f'(x) g(x) - f(x) g'(x)}{g^{2}(x)}$ , where f'(x) and g'(x) are the derivatives of each function. We use this to compute $\frac{\partial }{\partial x} \bigg(\frac{x}{b^{2} + r^{2}}\bigg) = \frac{x' (b^{2} + r^{2}) - (b^{2} + r^{2})' x}{(b^{2} + r^{2})^{2}} = \frac{(b^{2} + r^{2}) - (r^{2})' x}{(b^{2} + r^{2})^{2}} = \frac{(b^{2} + r^{2}) - 2r r' x}{(b^{2} + r^{2})^{2}}$ .

Because $r = \sqrt{x^2 + y^2 + z^2}$ , it's derivative respect to "x" is $r' = \frac{(x^2 + y^2 + z^2)'}{2\sqrt{x^2 + y^2 + z^2}} = \frac{2x}{2\sqrt{x^2 + y^2 + z^2}} = \frac{x}{\sqrt{x^2 + y^2 + z^2}}$

So, we continue with $2rr' = 2\sqrt{x^2 + y^2 + z^2}\frac{x}{\sqrt{x^2 + y^2 + z^2}} = 2x$ , and we put this in $\frac{\partial }{\partial x} \bigg(\frac{x}{b^{2} + r^{2}}\bigg) = \frac{(b^{2} + r^{2}) - 2r r' x}{(b^{2} + r^{2})^{2}} = \frac{(b^{2} + r^{2}) - 2xx}{(b^{2} + r^{2})^{2}} = \frac{b^{2} + r^{2} - 2x^2}{(b^{2} + r^{2})^{2}}$ . Since the other component are functionally the same expression, we can assure that $\frac{\partial }{\partial y} \bigg(\frac{y}{b^{2} + r^{2}}\bigg) = \frac{b^{2} + r^{2} - 2y^2}{(b^{2} + r^{2})^{2}}$ and $\frac{\partial }{\partial z} \bigg(\frac{z}{b^{2} + r^{2}}\bigg) = \frac{b^{2} + r^{2} - 2z^2}{(b^{2} + r^{2})^{2}}$ . To finish the problem we need to add all of the previous results to get $\nabla . \vec{E} = \frac{b^{2} + r^{2} - 2x^2}{(b^{2} + r^{2})^{2}}+ \frac{b^{2} + r^{2} - 2y^2}{(b^{2} + r^{2})^{2}} + \frac{b^{2} + r^{2} - 2z^2}{(b^{2} + r^{2})^{2}} = \frac{3b^{2} + 3r^{2} - 2x^2 - 2y^2- 2z^2}{(b^{2} + r^{2})^{2}}$