Physics Electric Charge and Field

When two or more individual charges are present in a system, the total charge will be an algebraic sum of all individual charges and not the vector sum. Therefore, an electric charge is considered as a scalar quantity.

This is an interesting question! Even though a proton has a positive charge, the net positive charge in a conducting material is always due to the removal of free electrons. 

This happens due to the availability of only free electrons in all conductors. Since protons are present in the nucleus, they can not roam freely in the conductor, as electrons travel freely in the conductor due to their presence in the outer shell.

Using gauss law in differential form

$\overrightarrow{\nabla }.\overrightarrow{E}=\frac{\rho }{{\epsilon }_0}$

$\overrightarrow{\nabla }.\left({\epsilon }_0\overrightarrow{E}\right)=\rho$

$\Rightarrow \overrightarrow{\nabla }.\overrightarrow{D}=\rho$

$\Rightarrow \left(\frac{\partial }{\partial x}\widehat{i}+\frac{\partial }{\partial y}\widehat{j}+\frac{\partial }{\partial z}\widehat{k}\right).\overrightarrow{D}=\rho$

$\begin{array}{l}\Rightarrow \rho =-e^{-x}siny+e^{-x}siny+2=2\\ \Rightarrow Q=2\left(Volume\right)=4*10^{-9}c=4nc\end{array}$

$F=\left[\frac{2k\lambda }{r_1}-\frac{2k\lambda }{r_2}\right]q=2k\lambda q\left[\frac{1}{r_1}-\frac{1}{r_2}\right]$

$2*9*10^9*3*10^{-6}\left[\frac{1000}{10}-\frac{1000}{12}\right]q$

$\begin{array}{l}\Rightarrow 4=9*10^{5}q\\ \Rightarrow q=4.44\mu C\end{array}$

In given circuit inductor behave as a simple wire so resultant circuit will be

R_ef = 2 + 1 = 3 $\Omega$  

V = IR

$l=\frac{30}{3}=10A$  

Direction of E in the direction of y axes so flux is only due to top and bottom surface for bottom surface y = 0 E = 0

and for top surface y = 0.5 m      So

$E=150*{\left(0.5\right)}^2=\frac{150}{4}$

$fluxflowing?=EA=\frac{150}{4}*{\left(0.5\right)}^2$              

$=\frac{150}{16}$             

Gausses law $?=\frac{q}{{\epsilon }_0}$  

  $\frac{150}{16}=\frac{q}{{\epsilon }_0}$            

$=8.3*10^{-11}C$              

When electric field is parallel then it would provide zero flux.

Viscous force = Weight

$\Rightarrow F_V=\rho *\frac{4}{3}\pi r^3*g=3.9*10N^{-10}$

dA = 2r dr

$dq=\sigma *2\pi rdr$

$\Rightarrow E={\displaystyle \underset{0}{\overset{R}{\int }}\frac{kdqz}{{\left(z^2+r^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}$

$\Rightarrow E=\frac{\sigma }{2{\epsilon }_0}\left[1-\frac{z}{\sqrt[]{R^2+z^2}}\right]$

(Independent of distance)

$E_1=E_2=\frac{\sigma }{2{\in }_0}$