Answer to Question #220736 in Electricity and Magnetism for Abhishek Gaurav

Question

https://i.postimg.cc/T2m6gKhG/20210726-175317.jpg

Questions is in photo

Answer 1

Gives

$\phi_1=xy^2z^2-3x^2y^2\\\phi_2=yx^2-6xy^2$

$E_1=-\nabla\phi_1\\E_2=-\nabla\phi_2$

$E_1=\nabla.(xy^2z^2-3x^2y^2)\\E_2=\nabla.(yx^2-6xy^2)$

$$E_1=(\frac{d}{dx}\hat{i}+\hat{j}\frac{d}{dy}+\hat{k}\frac{d}{dz})(xy^2z^2-3x^2y^2)$$

$$E_1=\hat{i}(y^2z^2-6xy^2)+\hat{j}(2xyz^2-6x^2y)+\hat{k}(xy^2)$$

$$E_2=(\frac{d}{dx}\hat{i}+\hat{j}\frac{d}{dy}+\hat{k}\frac{d}{dz})(yx^2-6xy^2)$$

$$E_2=\hat{i}(2xy-6y^2)+\hat{j}(x^2-12xy)+\hat{k}(0)$$