Electric potential is given by v = 6x – 8xy² – 8y + 6yz – 4z² then electric force acting on 2C point charge placed at origin will be

$\begin{array}{l}\mathrm{V}=6\mathrm{x}+8{\mathrm{xy}}^2-8\mathrm{y}+6\mathrm{yz}-4{\mathrm{z}}^2;\mathrm{Q}=2\mathrm{C}\\ \frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}=6-8{\mathrm{y}}^2;{\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}\right)}_{\mathrm{origin} }=6\\ \left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{y}}\right)=16\mathrm{xy}-8+6\mathrm{z};{\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}\right)}_{\mathrm{origin} }=-8\\ \frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{z}}=6\mathrm{y}-8\mathrm{z}\Rightarrow {\left(\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{z}}\right)}_{\mathrm{origin} }=0\\ \Rightarrow \overrightarrow{\mathrm{E}}=-\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{x}}\mathrm{i}-\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{y}}\mathrm{j}-\frac{\mathrm{\partial }\mathrm{V}}{\mathrm{\partial }\mathrm{z}}=-6\mathrm{i}+8\mathrm{j}\end{array}$
$\mathrm{E}=10{\mathrm{NC}}^{-1}\Rightarrow \mathrm{F}=\mathrm{EQ}=10\times 2=20\mathrm{N}$