37 - Electric Field and Potential Questions Answers

Asked By: TUSHAR GUPTA
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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

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

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

Potential difference from centre to surface = 3V/2 - V = V/2 here V = kq/r (q is charge of sphere) mvr2/2 = VQ/2 (Q is charge of particle) vr is radial velocity Besides tangential velocity due to rotation of sphere = rω=r2πn now solve  

Asked By: BHASKAR JHA
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Joshi sir comment

By volume comparision radius of combined drop = 21/3r net charge in the new drop = 2q now put these values in the formula of potential 

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

Since thickness is same so force per unit length should be same  Force per unit length = F/2πR  Charge density in the sphere is proportional to E so F α qE α σAE α E2R2 Thus force per unit length α E2R2/2πR α E2R Now compare the two cases

Asked By: KRISHAV RAJ SINGH
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Solution by Joshi sir

 

kq1q2l32=kq2q3l12=kq3q1l22 l1+l2+l3=L from these 2 equations  L 1q11q1+1q2+1q3,L 1q21q1+1q2+1q3,L 1q31q1+1q2+1q3 

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