69 - Electricity Questions Answers

Asked By: TUSHAR GUPTA
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Asked By: TUSHAR GUPTA
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Asked By: TUSHAR GUPTA
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Asked By: TUSHAR GUPTA
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Sir pls solve
Asked By: KANDUKURI ASHISH
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Joshi sir comment

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 charge2πxdxσ is V2πxdxσ put the value of V and integrate inform for any problem 

 

Asked By: SRIJAN
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Joshi sir comment

We know that planeEdScosθ = q6ε0   (1) and Net force on the plane =planeFdQ  =plane qEcosθdSσ =Qd2plane qEcosθdS   (2) Now solve eq. (1)and (2)

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

Total flux in 6 faces =σ4πr2ε0   so in 1 face = 16σ4πr2ε0  if any issue, inform

Sir how to solve? My main problem is how is current flowing between the spheres? Is all the current coming out from one sphere going to the other sphere? Sir can you show with a diagram where exactly all the current is going?  Thank you sir. 
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

rρdx4πx2=ρ4π-1x on putting limits resistance around  one sphere =k1r similar for other sphere so net R1=2kr In second case radius of one sphere  is half means r/2 Thus R2 = kr+kr/2=3kr now for same battery i1R1=i2R2 solve and wait for video    

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:

FinallyE = σ03ε0 = σ3ε0cosθ Now solve

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