Three small metal spheres AA, BB, and CC, each of radius rr, are supported on insulating stands and located at a distance 3a3a (where a≫ra≫r)

[rradmin]

 

[Amodini]

1. Electrostatic induction causes charges to rearrange on the conductors such that all spheres remain at the same potential.
2. The potential of each sphere is the sum of potentials due to all induced charges on A, B, and C.
3. Let induced charges be qA, qB, qC on A, B, C respectively.
4. Since they’re connected with a conductor:
   
   V_A = V_B = V_C

Let’s proceed to set up equations using the potential at each sphere due to:

• The point charge q
• The charges on spheres A, B, and C

We'll use:

Potential at a point due to charge Q at distance r=1/4πε0 ⋅ (Q/r)

Use symmetry:​

• The geometry is symmetric about B.
• Distances:
  • rAB = rBC = 3a
  • rAC = 6a

By symmetry of the arrangement and distances from point charge:

• Potentials from q on A and C are equal, and so the influence on A and C is symmetric.
• Therefore:
  
  q1 = q3

Let q1 = q3 = x, and q2 = y

 

[Abhineet]

Since the spheres are connected by a conducting wire, they will share charge and be at the same potential. The external charge q breaks the symmetry and induces charges on the spheres due to electrostatic induction.

Let the induced charges on spheres A, B, and C be q_A, q_B, and q_C. The total induced charge on all three spheres must be zero (as they were neutral initially and only affected by external charge q)

 

[Sneha Duhan]

 

[Faaiz]