Question

Verify that \({\mathop{\rm div}\nolimits} {\bf{E}} = 0\) for the electric field \({\bf{E}}({\bf{x}}) = \frac{{\varepsilon Q}}{{|{\bf{x}}{|^3}}}{\bf{x}}\).

Short answer

The gradient field of the function is:

\(\frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{i}} + \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{j}} + \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}}\).

Step-by-step solution

Statement of divergence theorem

The divergence theorem is a mathematical expression of the physical truth that, in the absence of matter creation or destruction, the density within a region of space can only be changed by allowing it to flow into or out of the region through its boundary.

Find the gradient field of the function.

Consider the given information.

\(f(x,y,z) = \sqrt {{x^2} + {y^2} + {z^2}} \)

Recall that,

\(\begin{array}{c}\nabla f = \frac{{\partial f}}{{\partial x}}(x,y,z){\bf{i}} + \frac{{\partial f}}{{\partial y}}(x,y,z){\bf{j}} + \frac{{\partial f}}{{\partial z}}(x,y,z){\bf{k}}\\ = \frac{{\frac{{\partial \left( {{x^2} + {y^2} + {z^2}} \right)}}{{\partial x}}}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{i}} + \frac{{2\frac{{\partial \left( {{x^2} + {y^2} + {z^2}} \right)}}{{\partial y}}}}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{j}} + \frac{{\frac{{\partial \left( {{x^2} + {y^2} + {z^2}} \right)}}{{\partial z}}}}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}}\\ = \frac{{2x}}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{i}} + \frac{{2y}}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{j}} + \frac{{2z}}{{2\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}}\\ = \frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{i}} + \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{j}} + \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}}\end{array}\)

Therefore, gradient field is \(\frac{x}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{i}} + \frac{y}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{j}} + \frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}{\bf{k}}\).