A non conducting ring of mass m and radius r is lying at rest in the vertcal XY plane on a smooth non conducting horizontal XZ plane. Charge +q & -q are distributed uniformly on the ring, on 2 sides of the vertical diameter of the ring.  A constant & uniform electric field E is set up along X direction.The ring is given a small rotation about the vertical diameter of the ring and released.Find the period of osscillation of the ring

 

Ans: 2 ^[2](mr/4qE)

 

The electric potential varies in space according to the relation V = 3x + 4y.A particle of mass 10 kg starts from rest from the point (2,3.2)under the influence of the field.Find the velocity of the particle when it crosses the X axis. The charge on the particle is + 1 C. Assume V and (x,y) are in SI Units.

 

Ans: 2 * 10 ^-3 m/s  

Please reply with detailed solution.

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THE QUESTION IS FROM ARIHANT ELECTRICITY AND MAGNETISM LEVEL 2 PROBLEMS BY D.C. PANDEY

 

1. getting T = 2pi sqrt ( mE / 8Eq )

 

please check the ans

 

2. easy  Ex = -3 N/kg

 

               Ey =  -4 N/kg

 

ay = q Ey/m

 

Now 3.2 = 1/2 ayt^2

 

find t.

 

calculate , vx = - ax t

 

  vy = - ay t

 

v = sqrt ( vx^2  + vy^2 )

 

ans = 2 . 10^-3  m/s