A pith ball of mass 9*10^(-5) kg carries a charge of 5 microcoulomb.What must be the charge of another pith ball placed 2 cm above given pith ball such that they are held in equilibrium

Question

experimentx · Answer

mg = kq1q2/r^2

anonymous · Answer

Why mg?

anonymous · Answer

isnt it Gm1m2/r^2=kq1q2/r^2

experimentx · Answer

9*10^(-5)*9.8 = 9x10^9 * 5x10^-6 * q2 /(2x10^-2)^2

experimentx · Answer

5 micro columb is quite huge amount of charge .. if it is in space ... where there is no gravity ... then your formula is correct.

anonymous · Answer

So gravity due to earth is counteracting electrostatic attraction?

experimentx · Answer

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anonymous · Answer

Oh I think I get it.I have been taking the mass of the other pith ball to be the same and equating those two equations.I see now.Thanx

experimentx · Answer

if this is the case ... on sphere on top of other