89 - Electricity Questions Answers

Let potential of a disc of radius x is V. For this, see video given below

Now potential energy of this disc with a circular layer of charge$$\begin{gathered} 2\mathrm{\pi xdx\sigma } \mathrm{is} \mathrm{V}2\mathrm{\pi xdx\sigma } \\ \mathrm{put} \mathrm{the} \mathrm{value} \mathrm{of} \mathrm{V} \mathrm{and} \mathrm{integrate} \\ \mathrm{inform} \mathrm{for} \mathrm{any} \mathrm{problem} \end{gathered}$$ 

$$\begin{gathered} We know that \\ \sum _{plane}EdS\cos \theta = \frac{q}{6{\epsilon }_0} \left(1\right) \\ and Net force on the plane =\sum _{plane}{F}_{\perp }dQ \\ =\sum _{plane} qE\cos \theta dS\sigma =\frac{Q}{d^2}\sum _{plane} qE\cos \theta dS \left(2\right) \\ Now solve eq. \left(1\right)\hspace{0.17em}and \left(2\right) \end{gathered}$$

$$\begin{gathered} Total flux in 6 faces =\frac{\sigma 4\pi r^2}{{\epsilon }_0} \\ so in 1 face = \frac{1}{6}\frac{\sigma 4\pi r^2}{{\epsilon }_0} \end{gathered}$$if any issue, inform

Current comes out from the surface of one sphere symmetrically 

and enters to the surface of other symmetrically.

Resistance of the medium around one sphere is

$$\begin{gathered} {\int }_r^{\infty }\frac{\rho dx}{4\pi x^2}=\frac{\rho }{4\pi }\left[-\frac{1}{x}\right] \\ on putting limits resis\tan ce around \\ one sphere =k\frac{1}{r} \\ similar for other sphere so net R_1=2\frac{k}{r} \\ In second case radius of one sphere \\ is half means r/2 \\ Thus R_2 = \frac{k}{r}+\frac{k}{r/2}=3\frac{k}{r} \\ now for same battery i_1{R}_1=i_2{R}_2 \\ solve and wait for video \end{gathered}$$ 

To nullify the electric field inside, the conducting sphere creates a situation in which electric field inside due to charge on sphere would be same as that of given electric field. Now this is the case, which is same as Irodov 3.17.

See the video now

link:

Finally$$\begin{gathered} E = \frac{{\sigma }_0}{3{\epsilon }_0} = \frac{\sigma }{3{\epsilon }_0\cos \theta } \\ Now solve \end{gathered}$$