33 - Electric Field and Potential Questions Answers

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.

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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 

A particle a having a charge of 5.0×10
5.0×10-7C
 is fixed in. a vertical wall. A second particle B of mass 100 g and. having equal charge is supended by a silk thread. of length 30 cm form the wall. The point of suspension is. 30 cm above the particle A. Find the angle of the thread. with the vertical when it stays in equilibrium.

Asked By: UDDESH KUMAR SABAT
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Solution by Joshi sir

A Volt-meter of 20000 Ohm resistance is connected with R ohm resistance in series. if we apply 110 V of pd to the mechanism, the Volt meter shows 5 V reading.. so find the R...

Asked By: BONEY HAVELIWALA
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Joshi sir comment

current on both the resistances will be same

so V1/R1 = V2/R2

5/20000 = 105/R

solve

An isolated conducting sphere of radius r has given charge q then PE??

ans is q2/8πe0r

But Sir ,why the ans q2/2πe0r is incorrect??

Asked By: SARIKA SHARMA
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Joshi sir comment

let we have given x charge and want to give dx also to the sphere

so dE = [1/4πε0]xdx/r

now integrate and take limits of x as 0 to q

2 Equal -ve charges -q r placed at pt(0,a) and (o,-a) on y axis , one +ve charge +q at rest is left from pt(2a,0). this charge will not execute SHM . WHY?? even when  F α -x condition is satisfying as force is due to both charges is acting & motion is also opposing during oscillating

Asked By: SARIKA SHARMA
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Joshi sir comment

Solve the question for force, you will get a condition according to which yours statement  "even when  F α -x condition is satisfying" 

is incorrect

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