Question

Three particles,each of mass ${10}^{-2}kg$ are brought from infinity to the vertices of an equilateral triangle of side 0.1 m. The minimum work that has to be done by external agent is :

• A. $-2\times {10}^{-8}J$
• B. $-2\times {10}^{-11}J$
• C. $-2\times {10}^{-12}J$
• D. $-2\times {10}^{-13}J$

Answer 1

Answer: (D)

$-2\times {10}^{-13}J$

Given,

$m_1=m_2=m_3=m\left(say\right)$

$m={10}^{-2}kg$

$r=0.1m$ side of equilateral triangle.

As, we know that the potential energy is scalar quantity,

$U=U_1+U_2+U_3$

$U=\frac{G{m}_1{m}_2}{r}+\frac{G{m}_2{m}_3}{r}+\frac{G{m}_3{m}_1}{r}$

$U=3\frac{Gmm}{r}$

$U=3\times \frac{6.67\times {10}^{-11}\times {10}^{-2}\times {10}^{-2}}{0.1}$

$U=20\times {10}^{-14}$

$U=2\times {10}^{-13}J$

The minimum work done is negative of potential energy,

$W=-U=-2\times {10}^{-13}J$

The correct option is D.