Physics - Electrostatics Question with Solution | TestHub

The electric field $\vec{E}$ is related to the electric potential $V$ by $\vec{E}=-\nabla V=-\left(\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\right)$. Given $V(x,y,z)=3x^2+4y^2-2z$. Partial derivatives are:$\frac{\partial V}{\partial x}=\frac{d}{dx}(3x^2)=6x$ $\frac{\partial V}{\partial y}=\frac{d}{dy}(4y^2)=8y$ $\frac{\partial V}{\partial z}=\frac{d}{dz}(-2z)=-2$ So, $\vec{E}=-(6x\hat{i}+8y\hat{j}-2\hat{k})=-6x\hat{i}-8y\hat{j}+2\hat{k}$. At the point $(1,1,1)\text{ m}$:$\vec{E}(1,1,1)=-6(1)\hat{i}-8(1)\hat{j}+2\hat{k}=-6\hat{i}-8\hat{j}+2\hat{k}\text{ V/m}$. The magnitude of the electric field is $|\vec{E}|=\sqrt{(-6{)}^2+(-8{)}^2+(2{)}^2}=\sqrt{36+64+4}=\sqrt{104}$.$\sqrt{104}\approx 10.198\text{ V/m}$. Rounding to the nearest integer, the magnitude of the electric field is $10\text{ V/m}$.