89 - Electricity Questions Answers
The bob of mass m , charge q is circulating in a vertical circle of radius R. With the help of a string . If the maximum speed of the bob is V then the period of revolution is T1. If an electric field of magnitude (mg/q) is setup which makes an angle 600 with upward vertical . again the bob is circulating in same circle and its maximum speed is also v , then the period of revolution is T2. Find then T1/T2.
in first case net force of gravity = mg
in second case electric force = qE = qmg/q = mg
so resultant of gravitational mg and electric mg will be mg again at an angle 60 degree with vertically downward direction.
in this time max velocity will not be at bottommost point as in previous case but it will make an angle 60 degree with horizontal
so T1 = T2
Consider the circuit shown with key opened. 3uf capacitor is charged to p.d. of 9 V. Capacitor 6 uF is charged to p.d. of 3V. The key is now closed . What will be the potential differnce across 3 uF capacitor in steady state?
V = [3*9-6*3]/(3+6) = 9/9 = 1
here - is used for opposite poles of the capacitors are connected together
Find the magnitude of electric potential at the origrn due to following charge distribution (if q = 1 nC)
sir i am unable to solve the series :
k - 3k/2 +5k/4 - 7k/4..........
= -k/2 - 3k/8 ....... plzz help
k - 3k/2 +5k/4 - 7k/8..........
= k(1-3/2+5/4-7/8+...............)
now let S = 1-3/2+5/4-7/8+.......
and -1/2 S = -1/2+3/4-5/8+......
on substracting 2nd by 1st
we get 3/2 S = 1-1+1/2-1/4+.......
or 3/2 S = 1-(1-1/2+1/4-1/8+..........)
now solve
A uniform electric field is established by connecting the plates of a parallel plate capacitor to a 12v battery. A charge particle of +6.24x10 to the power -6c moves from rest from the positive plate to the negative plate. when the charge reaches the negative plate,it acquires a speed of 3.4m/s.what is{a} the mass of the charge particle and{b} its final kinectic energy?
energy gained by the particle between the plates of the capacitor = qV
this will provide K.E. to the particle so qV = 1/2 mv2
solve for mass
then for K.E.
Calculate potential difference across a and b ,
since net potential = 0, answer will be 0 for the given diagram. By interchanging the poles of any one battery this question will become meannigfull so change it and apply Kirchhoff law.
a 24.6 mev alpha particle moves towards gold nucleus[atomic no 79] calculate the distance of closest approach of alpha particle
upto closest approach total KE of α particle will become PE
so use the formula KE = 9*109 * Ze2e/r
here KE = 24.6 mev convert it into joule first
solve equation for r
V = Ed = 6* 107*0.002
Find the potential difference Va - Vb between a and b ?
let charge in the line of 24 volt is Q in the direction of battery and at point b, let the charges in the two lines are q and Q-q (q in the middle line)
now accoring to the Kirchhoff law
24-12 = Q/1 + q/2
and 24-6 = Q/1 + (Q-q)/4
solve these 2 for q and Q
then by point potential rules get the answer.
two point charge are placed on x axis a 2mc charge at x=10 and -1 mc at x =40cm calculate the potential at pt x =100cm
potential at 100 cm = k 2 * 10-3 / 0.9 - k 1* 10-3 / 0.6
now solve
A POINT CHARGE q IS ROTATED ALONG A CIRCLE IN D ELECTRIC FIELD GENERATED BY ANOTHER POINT CHARGE Q D WORK DONE BY ELECTRIC FIELD ON ROTATING CHARGE IN ONE COMPLETE REVOLUTn IS ZERO
WHY???
Because this particle is revolving in an equipotential surface generated by other and work done = potential diff * q = 0