A metal sphere $B$ is initially charged with $+2 \, \mu \text{C}$ and is enclosed in the space between metal pieces $A$ and $C$ as shown. Then a charge of $+7 \, \mu \text{C}$ is transferred from metal $C$ to $A$. **Diagram Description:** - The diagram shows a metal sphere labeled $B$ with a charge of $+2 \, \mu \text{C}$. - Metal $A$ and metal $C$ are shown enclosing the sphere $B$. - An arrow indicates a transfer of $7 \, \mu \text{C}$ from metal $C$ to metal $A$.

Key Concepts Needed

1. Conservation of Charge
   The algebraic sum of all charges in a closed system remains constant; we can only move charge from one conductor to another.

2. Electrostatic Shielding in Conductors
   Inside the bulk of a conductor, the electric field is zero when static (electrostatic equilibrium). Any net charge must live on the outer surfaces of a conductor.

3. Gauss’s Law for Spherical Symmetry
   For a closed spherical Gaussian surface inside a conductor’s body, the enclosed charge must be zero. This forces induced charges to appear on the inner surfaces whenever some charge sits inside a hollow conductor.

Logical chain of thought to crack the problem

1. Initial Charges

   • Sphere B: $+2\,\mu\text{C}$ (given)
   • Assume large shells A and C start electrically neutral (that is how textbook examples are set up unless told otherwise).

2. Before transferring +7 \mu C

   • The +2 µC on B induces −2 µC on the inner surface of shell A (to kill the field inside A).
   • Therefore +2 µC must appear on A’s outer surface (so A stays overall neutral).
   • Shell C is still neutral and sees +2 µC sitting on A’s outside. To cancel field inside the metal of C, −2 µC is induced on C’s inner surface and +2 µC goes to C’s outer surface.

3. Transfer +7 µC from C to A

   • We physically take +7 µC off C’s outer surface and deposit it on A’s outer surface.
   • That action does not reach the inner surfaces directly, but after the move each conductor must re-adjust so that their interior fields vanish again.
   • Account for charges piece-by-piece:
     • Shell C loses +7 µC from its outer surface → C overall becomes −7 µC.
     • Shell A gains +7 µC on its outer surface → A overall becomes +7 µC.

4. Re-balance each conductor
   For each conductor, split the net charge between its inner and outer surfaces so that field inside the metal wall is zero:

   • Shell A (net +7 µC) already carries −2 µC on the inner surface (fixed by +2 µC of B). So the outer surface must hold
     $\text{Outer charge of A} = +7\,\mu\text{C} - (-2\,\mu\text{C}) = +9\,\mu\text{C}.$
   • Shell C (net −7 µC) already has −2 µC fixed on its inner surface (to cancel +9 µC of A seen inside). Therefore its outer surface must hold
     $\text{Outer charge of C} = -7\,\mu\text{C} - (-2\,\mu\text{C}) = -5\,\mu\text{C}.$

The sphere B stays +2 µC because it is insulated.

Why each step matters

• Induced charges on inner surfaces are dictated only by the charge enclosed inside that cavity (Gauss + zero field).
• Outer surface charge is simply the remainder once the inner requirement is satisfied.
• Conservation makes sure total charge moved equals +7 µC.

What is happening?

Imagine you have three hollow tin cans nested one after another.

• Tin-can B is a little ball in the middle, already holding 2 chocolate bars of positive charge ($+2\,\mu\text{C}$).
• Around B sits a bigger tin-can A.
• Around both of them there is an even bigger tin-can C.

Now somebody moves 7 extra chocolate bars of positive charge from the outer-most can (C) and sticks them on the middle can (A).

Because metal cans do not like electric field inside them, the charges will automatically rearrange themselves on their outer or inner skins so that the inside of every metal stays field-free. We must keep track of where each chocolate bar finally sits while the total number of chocolate bars never changes.

That is the puzzle you have to solve.

Step-by-Step Solution

1. Notation
   $q_B$ – charge on solid sphere B
   $q_{Ai}$, $q_{Ao}$ – charges on A’s inner and outer surfaces
   $q_{Ci}$, $q_{Co}$ – charges on C’s inner and outer surfaces

2. Initial arrangement (before +7 µC transfer)

   Sphere B:
   $q_B = +2\,\mu\text{C}$

   Inner surface of A (must cancel $q_B$):
   $q_{Ai} = -2\,\mu\text{C}$

   Shell A is neutral at start → outer surface has
   $q_{Ao} = +2\,\mu\text{C}$

   Charge seen inside C (all inside C is +2 µC) → inner surface of C:
   $q_{Ci} = -2\,\mu\text{C}$

   Shell C neutral → outer surface
   $q_{Co} = +2\,\mu\text{C}$

3. Transfer +7 µC from C to A
   Remove +7 µC from $q_{Co}$ and add it to $q_{Ao}$.

   After transfer:
   $q_{Co}^{\prime} = +2\,\mu\text{C} - 7\,\mu\text{C} = -5\,\mu\text{C}$
   $q_{Ao}^{\prime} = +2\,\mu\text{C} + 7\,\mu\text{C} = +9\,\mu\text{C}$

4. Check each conductor’s net charge

   Sphere B (isolated):
   $q_B = +2\,\mu\text{C} \quad (\text{unchanged})$

   Shell A:
   $q_{A,\text{net}} = q_{Ai} + q_{Ao}^{\prime} = -2\,\mu\text{C} + 9\,\mu\text{C} = +7\,\mu\text{C}$

   Shell C:
   $q_{C,\text{net}} = q_{Ci} + q_{Co}^{\prime} = -2\,\mu\text{C} + (-5\,\mu\text{C}) = -7\,\mu\text{C}$

   All good: total charge $+2+7-7 = +2 \mu\text{C}+$ (original charges) = +2 µC, then plus the +7 µC we moved (but total system started neutral for shells). Conservation satisfied.

5. Final answer (charges on surfaces)

   &\text{Sphere B} : +2\,\mu\text{C} \\ &\text{Inner surface of A} : -2\,\mu\text{C} \\ &\text{Outer surface of A} : +9\,\mu\text{C} \\ &\text{Inner surface of C} : -2\,\mu\text{C} \\ &\text{Outer surface of C} : -5\,\mu\text{C} \end{aligned}}$$