Four point masses easch of mass 'm' are placed at four vertieces A, B, C, nd D of a regular hexagon of side 'a' as shown in figure , Find gravitational potential and field strength at the centre O of the hexagon.
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Options

A. `-(4Gm)/(a) , (sqrt(3)(Gm))/(a^(2))`
B. `-(4Gm)/(a) , (sqrt(3)(Gm))/2(a^(2))`
C. `-(Gm)/(a) , (sqrt(3)(Gm))/(a^(2))`
D. `-(2Gm)/(a) , (sqrt(3)(Gm))/(a^(2))`

Answer

A, B, C

Explanation

Gravitational potential id scalar quantity. Therefore,
`V_0 =`scalar sul of gravtational potentials produced by four point massses at A, B, C, and D.
`= (Gm)/(a)-(Gm)/(a)-(Gm)/(a)`
`= -(4Gm)/(a)`
gravtational field strength is a vector quantity. So, it is a vector sum of four vectors of equal magnitudes.
`E = (Gm)/(a^2) = E_A = E_B = F_C = E_D`

`E_A` and `E_D` are cancelled. So, net field dtrength is a vector sum of `E_B` and `E_C` at angle `60@`.
`:. E_(net) = sqrt(E^2+E^2+2(E)(E)cos60^(@)) = sqrt(3)E`
`=(sqrt(3)(Gm))/(a^(2))`
this net field strength is along the bisector line of `/_COB`, away from O, between `E_(C)` and `E_(B)`.