51 - Electric Field and Potential Questions Answers
first calculate the magnitude of electric field at centre of ring due to charge on the surface of ring and direction of that electric field also
this can be solved by method of integration
then F = qE
if uncleared reply, we will give complete answer
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.
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
TWO EQ. +VE CHARGES R KEPT AT PT A & B D ELECTRIC POTENTIAL AT D PT B/W A & B (EXCLUDING THESE POINTS ) IS STUDIED WHILE ,MOVING FROM A TO B D POTENTIAL DECREASE THEN INCREASE WHY???
at start when the point will be nearer to first charge, potential due to the nearer charge will be very high, at the middle point of the two charges it will be minimum and after that it will increase again due to its approach towards another charge.
There are two fixed spheres of charges Q and 4Q and the distance between their centres is 6R. The Radius of first sphere is R and radius of the secind one is 2R. A charge -q is to be projected from the along the line joining their centres ( it is on the surface of the sphere with radius R), so that it can reach the surface of other solid sphere of charge 4Q. What will be the minimum velocity to be given so that it reaches there?
sir i have used energy conservation in this problem :
-kqQ/R - kQ4q/5R +.5 mv^2 = -kQq/4r - kqQ4/2R
but im not getting the answer .....is anything left in ths ..please telll
first find a point midway between the two spheres where net E will be 0
Then consider v = 0 at that point and compare the potential energy + kinetic energy of the starting point and that obtained point.