Answer to Question #106510 in Electricity and Magnetism for lolipop

Question

[img]https://upload.cc/i1/2020/03/25/EN0FM7.jpg[/img]



the question is in the picture,R=4

Answer 1

As per the given question,

$\overrightarrow{E}=(R+2)\hat{i}+5\hat{j}$

$\overrightarrow{dA}=dA(\hat{i}+3\hat{j})$

R=4

$\oint dA=2m^2$

We know, the magnetic flux

$\phi=\oint\overrightarrow{E}.\overrightarrow{A}=\dfrac{q}{\epsilon_o}$

$\Rightarrow \phi=\oint((R+2)\hat{i}+5\hat{j})(\hat{i}+3\hat{j})dA$

$\Rightarrow\phi=(R+2)\oint dA+15\oint dA$

Now, substituting the values,

$\phi=(R+2)2+15\times2 =2R+4+30$

$\Rightarrow\phi=2R+34$

So, $\phi=2\times 4+34$

$\Rightarrow \phi=42 N-m^2/C$

now, $\phi= \dfrac{Q}{\epsilon_o}$

$\dfrac{Q}{\epsilon_o}=42$

$\Rightarrow Q=42\epsilon_o= 371.7 \times 10^{-12}C$