Answer to Question #106256 in Electricity and Magnetism for lolipop

Question

[img]https://upload.cc/i1/2020/03/23/PHdlM4.jpg[/img]


the question is in the picture ,R =4

Answer 1

Question

Given $\vec{E}=(R+2)\vec{i}+5\vec{j}\text{ N/C}, \space\text{d}\vec{A}=\text{d}A(\vec{i}+3\vec{j}),$ and $\oint\text{d}A=2\text{ m}^2,$ draw the Gaussian surface in figure 2 below that satisfies flux $\Phi$ of the given condition. Show all your calculation steps clearly. Correct your final answer to the nearest 10 of pC 892.5 pc = 890 pC. $R=4.$

Solution

What is flux?

$$\Phi=\oint \vec{E}\cdot\text{d}\vec{A}=\\ =(6\hat{i}+5\hat{j})\cdot(\hat{i}+3\hat{j})=\\ =(6+15)\cdot2=42\space\frac{\text{Nm}^2}{\text{C}}.$$

On the other hand, the the Gaussian surface can be expressed as:

$$\Phi=\frac{\Sigma q_\text{enclosed}}{\epsilon_0},\\ \space\\ \Sigma q_\text{enclosed}=\Phi\epsilon_0=370\text{ pC}.$$

Therefore, now we must draw the Gaussian surface that encloses exactly 370 pC: