Four identical point masses, each of mass m and carrying charge +q are placed at the corners of a square of sides 'a' on a frictionless plain surface. If the particles are released simultaneously, the kinetic energy of the system when they are infinitely far apart is

Question

Four identical point masses, each of mass m and carrying charge +q are placed at the corners of a square of sides 'a' on a frictionless plain surface. If the particles are released simultaneously, the kinetic energy of the system when they are infinitely far apart is (A)  (q2/a)(2 √2+1) (B) (q2/a)(√2+2) (C) (q2/a)(√2+4) (D) (q2/a)(√2+1)

Tardigrade Admin · Accepted Answer

K Ei+P Ei=K Ef+P Ef ∴ K Ei+P Ei=4 (K q2/a)+2 (K q2/√2 a) =(K q2/a)(4+√2), K =1 in C.G.S. unit

Q. Four identical point masses, each of mass $m$ and carrying charge $+ q$ are placed at the corners of a square of sides ' $a$ ' on a frictionless plain surface. If the particles are released simultaneously, the kinetic energy of the system when they are infinitely far apart is

$\frac{q ^{2}}{a} (22 + 1)$

$\frac{q ^{2}}{a} (2 + 2)$

$\frac{q ^{2}}{a} (2 + 4)$

$\frac{q ^{2}}{a} (2 + 1)$

$K E_{i} + P E_{i} = K E_{f} + P E_{f}$
$∴ K E_{i} + P E_{i} = 4 \frac{K q ^{2}}{a} + 2 \frac{K q ^{2}}{2 a}$
$= \frac{K q ^{2}}{a} (4 + 2), K = 1$ in C.G.S. unit

Q. Four identical point masses, each of mass m and carrying charge +q are placed at the corners of a square of sides 'a' on a frictionless plain surface. If the particles are released simultaneously, the kinetic energy of the system when they are infinitely far apart is

aq2​(22​+1)

aq2​(2​+2)

aq2​(2​+4)

aq2​(2​+1)