# 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.

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

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

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...

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??

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