# 37 - Electric Field and Potential 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

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

a small ball of mass 1kg and a charge 2/3 uC is placed at the center of a uniformly charged sphere of radius 1m and charge 1/3 mC. a narrow smooth horizontal groove is made in the sphere from centre to surface as shown in figure. the sphere is made to rotate about its vertical diameter at a constant rate of 1/2pi revolutions per second. find the spedd w.r.t ground with which the ball slide out from the groove. neglect any magnetic force acting on ball?

**Asked By: VISHNU VARDHAN**

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**Joshi sir comment**

$Potentialdifferencefromcentretosurface=3V/2-V=V/2\phantom{\rule{0ex}{0ex}}hereV=kq/r(qischargeofsphere)\phantom{\rule{0ex}{0ex}}m{{v}_{r}}^{2}/2=VQ/2(Qischargeofparticle)\phantom{\rule{0ex}{0ex}}{v}_{r}isradialvelocity\phantom{\rule{0ex}{0ex}}Besides\mathrm{tan}gentialvelocityduetorotationofsphere=r\omega =r2\pi n\phantom{\rule{0ex}{0ex}}nowsolve\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

**Asked By: BHASKAR JHA**

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**Joshi sir comment**

$Byvolumecomparisionradiusofcombineddrop={2}^{1/3}r\phantom{\rule{0ex}{0ex}}netchargeinthenewdrop=2q\phantom{\rule{0ex}{0ex}}nowputthesevaluesintheformulaofpotential\phantom{\rule{0ex}{0ex}}$

Initially the spheres a and b are at potential be a and b respectively now sphere b is earthd by closing the switch the poential of a will become

**Asked By: ROMEO ROMI**

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**Joshi sir comment**

No meaning can be incurred by this language. Rectify it for getting answer

**Asked By: DARIUS**

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**Joshi sir comment**

$Sincethicknessissamesoforceperunitlengthshouldbesame\phantom{\rule{0ex}{0ex}}Forceperunitlength=F/2\pi R\phantom{\rule{0ex}{0ex}}ChargedensityinthesphereisproportionaltoE\phantom{\rule{0ex}{0ex}}soF\alpha qE\alpha \sigma AE\alpha {E}^{2}{R}^{2}\phantom{\rule{0ex}{0ex}}Thusforceperunitlength\alpha {E}^{2}{R}^{2}/2\pi R\alpha {E}^{2}R\phantom{\rule{0ex}{0ex}}Nowcomparethetwocases$

**Asked By: KRISHAV RAJ SINGH**

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**Solution by Joshi sir**

$\frac{k{q}_{1}{q}_{2}}{{{l}_{3}}^{2}}=\frac{k{q}_{2}{q}_{3}}{{{l}_{1}}^{2}}=\frac{k{q}_{3}{q}_{1}}{{{l}_{2}}^{2}}\phantom{\rule{0ex}{0ex}}{l}_{1}+{l}_{2}+{l}_{3}=L\phantom{\rule{0ex}{0ex}}fromthese2equations\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{L\frac{1}{\sqrt{{q}_{1}}}}{\frac{1}{\sqrt{{q}_{1}}}+\frac{1}{\sqrt{{q}_{2}}}+\frac{1}{\sqrt{{q}_{3}}}},\frac{L\frac{1}{\sqrt{{q}_{2}}}}{\frac{1}{\sqrt{{q}_{1}}}+\frac{1}{\sqrt{{q}_{2}}}+\frac{1}{\sqrt{{q}_{3}}}},\frac{L\frac{1}{\sqrt{{q}_{3}}}}{\frac{1}{\sqrt{{q}_{1}}}+\frac{1}{\sqrt{{q}_{2}}}+\frac{1}{\sqrt{{q}_{3}}}}\phantom{\rule{0ex}{0ex}}$