Answer to Question #193671 in Electricity and Magnetism for laura

Question

consider the vector A=sin(π÷2)i - sin(π÷2x)j, what is the divergence of the vector A

Answer 1

Gives

$A=sin\frac{\pi}{2} \hat{i}+sin\frac{\pi}{2x}\hat{j}$

Where$sin\frac{\pi}{2}=1$

$\nabla. A=(\frac{d}{dx}\hat{i}+\frac{d}{dy}\hat{j}).(sin\frac{\pi}{2} \hat{i}+sin\frac{\pi}{2x}\hat{j})$

$\nabla. A=\frac{d}{dx}(1)+\frac{d}{dy}(sin\frac{\pi}{2x})$$\nabla.A=0+0=0$

$\nabla .A=0$