Question

Calculate work done to dissociate the system of three charges placed on the vertices of a triangle as shown.
Here q = 1.6X10-10 C.

Answer 1

Let Q₁ , Q₂ , Q₃ charges are placed on vertices A , B , C respectively.
See attachment, Q₁ = q , Q₂ = -4q and Q₃ = +2q

Now, potential energy of system of three charged particles is given by
U = [KQ₁Q₂/r₁ + KQ₂Q₃/r₂ + KQ₃Q₁/r₃ ]
For equilateral triangle, r₁ = r₂ = r₃ = r = 10 cm[ given ]

Now, U = [Kq(-4q)/r + K(-4q)(+2q)/r + K(+2q)(q)/r ]
= [ -4Kq²/r -8Kq²/r + 2Kq²/r ]
= -10Kq²/r

Now, put values of q and r
U = -10 × 9 × 10⁹ × (1.6 × 10⁻¹⁰)²/(0.1)
= 9 × 2.56 × 10⁻⁹ J = -2.304 × 10⁻⁸ J

Now, work done = negative of potential energy
= -(-2.304 × 10⁻⁸)J
= 2.304 × 10⁸ J