Electrostatics> a charged particle having some mass is re...

a charged particle having some mass is resting in equilibrium at height h above the centre of a uniformly charged non conduction horizontal ring of radius R, The force of gravity acts downwards , the equilibrium of the particle will be stable : for all values of h only if h>R/root2 only if h only ifh=R/root2

Mansi , 8 Years ago

Grade 12

2 Answers

Vikas TU

Last Activity: 8 Years ago

Technically, the correct answer is 'none of the above'. If you look up Earnshaw's theorem, you can find out why.

However if the mass is constrained so it can't move sideways (e.g. it is a bead on a vertical wire) then the problem can be solved.

The field along the axis of the ring reaches a maximum value when H = R/sqrt(2). For a proof of this, see the link.

For the particle to be stable:
a) a small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it; so the electric force must decrease if H is increased from the equilibrium position

b) a small vertical displacement downwards should cause resultant force on the particle to be upwards, to return it; so the electric field must increase if H is decreased from the equilibrium position.

The conditions can only be true if H > R/sq.root2 , because in this region the electric field increase if H decreases and decreases of H increases.

Approved

Kushagra Madhukar

Last Activity: 4 Years ago

Dear student,

Please find the answer to your question below.

The conditions for stable equilibrium can only be true if H > R/√2 , because in this region the electric field increase if H decreases and decreases if H increases. Due to which a slight displacement in vertical direction either upwards or downwards brings back the charge particle back to the same place eventually or the net resultant force acts always towards the direction of equilibrium position.

Thanks and regards,

Kushagra