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Two equal negative charges q are fixed at points (0,a) and (0,a) on yaxis. A positive charge Q is released from rest at the point (2a,0) on the xaxis. The charge Q will
Positive and negative point charges of equal magnitude are kept at (0,0,a/2) and (0,0,a/2), respectively. The work done by the electric field when another positive point charge in moved from (a,0,0) to (0,a,0) is.
An electron of mass me , initially at rest, moves through a certain distance in a uniform electric field in time t1 .A proton of mass mp ,also initially at rest, takes time t2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t1 /t2 is nearly equal to.
Answer : c
Solution : If the acceleration is a then the distance s travelled in time t is given by
S = 1/2at^2
Therefore for electron
S = ½ a(e)t2^1
And for Electron
S = ½ a(p)t2^2
Thus, t2/t1 = [a(e)/a(p)]^1/2.......... (1)
Force on electron in an electric field E
F(e) = eE
Thus the acceleration of the electron if its mass is
a(e) = F(e)/Me => eE/Me
Similarly the acceleration of the proton
a(p) = F(e)/Mp => eE/Mp
Therefore, a(e)/a(p) = Mp/Me
Therefore, from (1)
t2/t1 =(Mp/Me)^1/2
Two charges, each equal to q, are kept at x=a and x=a on the xaxis. A particle of mass m and charge q_{0}=q/2 is placed at the origin. If charge q_{0 }is given a small displacement y(y<<a) along the yaxis, the net force acting on the particle is proportional to
A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium if q is equal to
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