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Find the potential energy of a system of four particles placed at the vertices of a square of side 1. Also obtain the potential at the centre of the square. |
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Answer» Solution :Consider four masses each of mass m at the corners of a square of side 1. see fig. we have four mass pairs at DISTANCE 1 and two DIAGONAL pairs at distance `SQRT(2)l` HENCE, `U(r) = -(4)(Gm^(2))/(1) - 2(Gm^(2))/(sqrt(2)l)` `= -(2Gm^(2))/(1)(2+(1)/(sqrt(2))) = -5.41(Gm^(2))/(1)` The gravitational potential at the centre of the square `(r = sqrt(2)l//2)` is `V(r) = -4sqrt(2)(Gm)/(1)`
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