two small equal point charges of magnitude q are suspended from a common point on the ceiling by insulating mass less strings of equal lengths.they come to equilibrium with each string making an angle "theta" with the vertical.if mass of each charge is m ,then the electrostatic potential at the center of line joining them will be what??

Asked by prabhakar p | 14th May, 2013, 07:30: AM

Expert Answer:

On the point charge, there will be 3 forces acting
1. T along the string
2. mg in the downward vertical direction
3. Electrostatic force in the horizontal direction away from the other point charge = kq^2/r^2
Since the point charge is in equilibrium
Hence, Tcos(theta) = mg
Also Tsin(theta) = kq^2/r^2
mgtan(theta) = kq^2/r^2
Hence, r = sqrt( kq^2/mgtan(theta))
Now, electrostatic potential at the mid point of the line joining 2 charges  (V) = kq/(r/2) + kq/(r/2)
V = 4kq/r
V = 4kq/sqrt( kq^2/mgtan(theta))
V = 4sqrt(kmgtan(theta))
Hence, option 4 is correct. 

Answered by  | 15th May, 2013, 08:58: PM

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