Question

Four particles each of mass m are placed at the vertices of a square of side l. the potential at the centre of square is
A. `(4sqrt(2)Gm)/(L)`
B. `-(4sqrt(2)Gm)/(L)`
C. zero
D. `-(8sqrt(2)Gm)/(L^(2))`

Answer 1

Correct Answer - B

`BD^(2)=L^(2)+L^(2)" " therefore BD=sqrt(2)L`
`therefore AO=BO=CO=DO=(sqrt(2)L)/(2)=(L)/(sqrt(2))`
Potential at O due to the mass at `A=-(Gm)/(AO)=-(Gm)/(L//sqrt(2))`
`therefore` Potential at O due to all the four particles
`V=-((4Gm)/(L//sqrt(2)))=(4sqrt(2)Gm)/(L)`