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.

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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Finally

Joshi sir comment

By superposition principle, it is solved in minimum time and effort, other method is very complex. So use superposition principle.

Divide current in point 1 and 2, considering symmetry throughout, we get the equivalent resistance of the whole network except resistor R between 1 and 2 is 2R

Now 2R and R in parallel between 1 and 2

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?

Joshi sir comment

Joshi sir comment

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

Joshi sir comment

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

Joshi sir comment

Solution by Joshi sir

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.

Solution by Joshi sir

if an area of isoscale right angle triangle(made of uniform wire of unit resistivity) is 8 unit. a battery of emf 10 unit is connected with across the longest branch then the current drawn from battery is approx                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                     ans is  3A

Joshi sir comment

area = l2/2 = 8 so l = 4

ρ = 1

so  resistance of the 3 legs of the triangle = 4/A, 4/A, 4√2/A

battery is connected across the longest branch so eq. resistance = 4/A+4/A+4√2/A = 13.6/A

i think area of cross section will be given then calculate current

Can You explain some easy way to find potential between A and B?

Joshi sir comment

use kirchhoff laws

4-10i+1-30x = 0

1-30x+20(i-x) = 0

solve it for i and x

potential across A and B = 4-10i+1 = 30x