The potential energy of a system of four particles placed at the vertices of a square of side l is

Question

The potential energy of a system of four particles placed at the vertices of a square of side l is (A) -3.71 (G m2/l) (B) -4.51 (G m2/l) (C) -5.41 (G m2/l) (D) -6.45 (G m2/l)

Tardigrade Admin · Accepted Answer

W=-4((G m2/l))-2((G m2/l √2)) ⇒ W=-(2 G m2/l)(2+(1/√2)) =-5.41((G m2/l)) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/15094c228cd45333478ab204a56386a8-.png" />

Q. The potential energy of a system of four particles placed at the vertices of a square of side $l$ is

$- 3.71 \frac{G m ^{2}}{l}$

$- 4.51 \frac{G m ^{2}}{l}$

$- 5.41 \frac{G m ^{2}}{l}$

$- 6.45 \frac{G m ^{2}}{l}$

$W = - 4 (\frac{G m ^{2}}{l}) - 2 (\frac{G m ^{2}}{l 2})$
$\Rightarrow W = - \frac{2 G m ^{2}}{l} (2 + \frac{1}{2})$
$= - 5.41 (\frac{G m ^{2}}{l})$

Q. The potential energy of a system of four particles placed at the vertices of a square of side l is

−3.71lGm2​

−4.51lGm2​

−5.41lGm2​

−6.45lGm2​