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4. A charged ball is launched between the horizontal plates of a large capacitor, along the middle plane, 1 meter from each plate. The ball passes undeflected through the capacitor. Now the field direction in the capacitor is reversed, and the charged ball launched again. How long in seconds will it travel before colliding with the lower plate of the capacitor? Draw a picture of the before and after field reversal launches, showing clearly all forces, as well as the electric field lines. Determine the charge to mass ratio of the ball, in coul/kg, if the voltage between the plates is 100 kV
5. Estimate the maximum voltage generated when a vertically mounted copper hoop is continuously rotated at 1000 rpm about an axis through its center, perpendicular to the earth's magnetic field. The hoop has a radius of 0.5 m, and is of uniform circular cross section, of diameter 1 cm. Find the maximum current in the hoop. Assuming all ohmic heating is conductive (no radiative losses), estimate how long it will be before the loop reaches melting point. Take standard Earth's magnetic field. Estimate the initial torque it takes to power the rotation if it takes 10 seconds for the hoop to reach full rotational speed
6.A charged bumblebee of mass M flies at a constant speed vo near a high tension wire carrying a current I. Prove, using the right hand rules, that if it flies at a certain distance d from the wire, in a certain direction, it would feel sustained periods of weightlessness. Determine its charge Q in terms of the other symbolic quantities. Draw a picture showing the relevant aspects of the situation. (Neglect the Earth's magnetic field.) How might the bee have ended up with a charge?
Comments (3)
say the current moving from L to R then by right hand rule B out of the plane and q(vxB) in upward direction so v has to be from R to L
so bee has to fly opposite to current
now due to wire B = (mu I) / (2 pi d)
force =(q v mu I) /(2 pi d) = mg =====> q = (2 pi d mg) / (v mu I)
how the bee got the charge
if the bee is near the wire the bee + wire system will have some capacitance so some charge will be induced on bee but net charge will remain zero. now if bee sits on some grounded object in the vicinity then it will be at zero potential but since initially net charge was zero finally it will have some non zero charge
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say q is charge on sphere and Q on capacitor then initially -Q has to be on top plate and +Q on bottom
equating electric force with gravity
now second time both electric and gravitational forces act downward
so net force is 2mg downward and acceleration is 2g
so 1 = gt^2 =====> t = 1/root(g)
middle plane at a distance 1 m from each plate so distance d = 2 m
since E const E*2 = 100 * 10^3 ===> E = 5 * 10^4
Eq = mg
q/m = g/E = 2*10^-4