# 69 - Electricity Questions Answers

Find potential energy of disc having charge density sigma uniformly distributed on its surface,radius R

**Asked By: AKASH MAURYA**

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**Joshi sir comment**

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$2\mathrm{\pi xdx\sigma}\mathrm{is}\mathrm{V}2\mathrm{\pi xdx\sigma}\phantom{\rule{0ex}{0ex}}\mathrm{put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{V}\mathrm{and}\mathrm{integrate}\phantom{\rule{0ex}{0ex}}\mathrm{inform}\mathrm{for}\mathrm{any}\mathrm{problem}$

**Asked By: SRIJAN**

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**Joshi sir comment**

$Weknowthat\phantom{\rule{0ex}{0ex}}\sum _{plane}EdS\mathrm{cos}\theta =\frac{q}{6{\epsilon}_{0}}\left(1\right)\phantom{\rule{0ex}{0ex}}andNetforceontheplane=\sum _{plane}{F}_{\perp}dQ\phantom{\rule{0ex}{0ex}}=\sum _{plane}qE\mathrm{cos}\theta dS\sigma =\frac{Q}{{d}^{2}}\sum _{plane}qE\mathrm{cos}\theta dS\left(2\right)\phantom{\rule{0ex}{0ex}}Nowsolveeq.\left(1\right)\hspace{0.17em}and\left(2\right)$

A hollow spherical shell of radius r has a uniform charge density sigma. It is kept in a cube of edge 3r such that the center of the cube coincides with the centre of the shell. The electric flux coming out of a face of the cube is

Sir pls sove

**Asked By: KANDUKURI ASHISH**

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**Joshi sir comment**

$Totalfluxin6faces=\frac{\sigma 4\pi {r}^{2}}{{\epsilon}_{0}}\phantom{\rule{0ex}{0ex}}soin1face=\frac{1}{6}\frac{\sigma 4\pi {r}^{2}}{{\epsilon}_{0}}$if any issue, inform

**Asked By: DARIUS**

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**Joshi sir comment**

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

${\int}_{r}^{\infty}\frac{\rho dx}{4\pi {x}^{2}}=\frac{\rho}{4\pi}\left[-\frac{1}{x}\right]\phantom{\rule{0ex}{0ex}}onputtinglimitsresis\mathrm{tan}cearound\phantom{\rule{0ex}{0ex}}onesphere=k\frac{1}{r}\phantom{\rule{0ex}{0ex}}similarforotherspheresonet{R}_{1}=2\frac{k}{r}\phantom{\rule{0ex}{0ex}}Insecondcaseradiusofonesphere\phantom{\rule{0ex}{0ex}}ishalfmeansr/2\phantom{\rule{0ex}{0ex}}Thus{R}_{2}=\frac{k}{r}+\frac{k}{r/2}=3\frac{k}{r}\phantom{\rule{0ex}{0ex}}nowforsamebattery{i}_{1}{R}_{1}={i}_{2}{R}_{2}\phantom{\rule{0ex}{0ex}}solveandwaitforvideo\phantom{\rule{0ex}{0ex}}$

A solid conducting sphere of radius R is placed in a uniform electric field E as shown in figure-1.441. Due to electric field non uniform surface charges are induced on the surface of the sphere. Consider a point A on the surface of sphere at a polar angle θ from the direction of electric field as shown in figure. Find the surface density of induced charges at point A in terms of electric field and polar angle θ.

**Asked By: DEBRAJ SAHA**

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**Joshi sir comment**

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$E=\frac{{\sigma}_{0}}{3{\epsilon}_{0}}=\frac{\sigma}{3{\epsilon}_{0}\mathrm{cos}\theta}\phantom{\rule{0ex}{0ex}}Nowsolve$