Challenge Solution: Charge Distribution on Spheres
Important Note on Labels:
This solution strictly follows the diagram provided below where Sphere B is at the 90° corner.
(Note: The original problem text implied “A” might be the corner in its answer key, but we are solving based on the visual diagram provided: B at corner, A and C at acute angles).
Figure 1: Configuration of spheres. B is at the right angle.
1. Problem Setup
This solution strictly follows the diagram provided below where Sphere B is at the 90° corner.
(Note: The original problem text implied “A” might be the corner in its answer key, but we are solving based on the visual diagram provided: B at corner, A and C at acute angles).
Figure 1: Configuration of spheres. B is at the right angle.
We have three identical metal spheres of radius \(r\) with initial charge \(q\). The distances are:
- \(d_{AB} = a\)
- \(d_{BC} = a\)
- \(d_{AC} = a\sqrt{2}\) (Hypotenuse)
Condition: \(r \ll a\). This means the potential of a sphere is dominated by its self-charge (\(V \approx kQ/r\)), but charge flow is driven by the small potential differences caused by neighbors.
Let the “Interaction Factor” be defined as: $$ \Lambda = \frac{2-\sqrt{2}}{4} = \frac{1}{2}\left(1 – \frac{1}{\sqrt{2}}\right) $$ Let \(\xi\) be a charge transfer unit: $$ \xi = \frac{rq}{a} \Lambda $$
2. Step-by-Step Derivation
Step 1: Connect A and B
Initial State: \(Q_A=q, Q_B=q, Q_C=q\).
We connect A (Top) and B (Corner). Charge flows until potentials equalize.
$$V_A = \frac{kQ_A’}{r} + \frac{kQ_B’}{a} + \frac{kQ_C}{a\sqrt{2}}$$
$$V_B = \frac{kQ_B’}{r} + \frac{kQ_A’}{a} + \frac{kQ_C}{a}$$
Note that \(V_B\) has a higher external potential than \(V_A\) because B’s neighbor C is closer (\(a\)) than A’s neighbor C (\(a\sqrt{2}\)). Thus, charge flows from High Potential (B) to Low Potential (A).
Equating \(V_A = V_B\) and solving for charge difference:
$$ \frac{Q_A’ – Q_B’}{r} \approx \frac{q}{a} – \frac{q}{a\sqrt{2}} = \frac{q}{a}\left(1 – \frac{1}{\sqrt{2}}\right) $$ $$ Q_A’ – Q_B’ = \frac{rq}{a}\left(1 – \frac{1}{\sqrt{2}}\right) = 2 \cdot \frac{rq}{a} \Lambda = 2\xi $$Since total charge \(2q\) is conserved:
- \(Q_A’ = q + \xi\) (Gained charge)
- \(Q_B’ = q – \xi\) (Lost charge)
- \(Q_C’ = q\) (Unchanged)
Step 2: Connect B and C
Current State: \(Q_A=q+\xi, \quad Q_B=q-\xi, \quad Q_C=q\).
We connect B (Corner) and C (Right). Potentials equalize.
$$V_B = \frac{kQ_B”}{r} + V_{ext, B} \quad \text{and} \quad V_C = \frac{kQ_C”}{r} + V_{ext, C}$$
There are two driving forces here:
- Self-Charge Difference: \(Q_C > Q_B\) implies \(V_C > V_B\). This drives charge from C &to; B.
- External Potential (from A): A is distance \(a\) from B, and distance \(a\sqrt{2}\) from C. So A raises B’s potential more than C’s. This drives charge from B &to; C.
Let’s find the magnitude. The condition \(V_B = V_C\) gives:
$$ \frac{Q_B” – Q_C”}{r} = \frac{Q_A}{a\sqrt{2}} – \frac{Q_A}{a} = -Q_A \left( \frac{1}{a} – \frac{1}{a\sqrt{2}} \right) $$Approximating \(Q_A \approx q\):
$$ Q_B” – Q_C” \approx – \frac{rq}{a} \left( 1 – \frac{1}{\sqrt{2}} \right) = -2\xi $$This means \(Q_C”\) must be larger than \(Q_B”\) by \(2\xi\).
Total charge involved: \(Q_B + Q_C = (q-\xi) + q = 2q – \xi\).
Solving the system \(\begin{cases} Q_B” + Q_C” = 2q – \xi \\ Q_C” – Q_B” = 2\xi \end{cases}\)
- \(Q_B” = q – 1.5\xi\)
- \(Q_C” = q + 0.5\xi\)
Step 3: Connect C and A
Current State: \(Q_A=q+\xi, \quad Q_B=q-1.5\xi, \quad Q_C=q+0.5\xi\).
We connect C (Right) and A (Top). Potentials equalize.
Symmetry Check: Sphere B is the remaining sphere. Distance \(d_{BA} = a\) and distance \(d_{BC} = a\).
Since B is equidistant from A and C, it exerts the same potential on both.
Therefore, the condition \(V_A = V_C\) simplifies entirely to their self-charges:
$$ \frac{kQ_A”’}{r} + V_{from B} = \frac{kQ_C”’}{r} + V_{from B} \implies Q_A”’ = Q_C”’ $$They simply share the total available charge equally.
$$ Q_{total} = Q_A + Q_C = (q + \xi) + (q + 0.5\xi) = 2q + 1.5\xi $$ $$ Q_A”’ = Q_C”’ = \frac{2q + 1.5\xi}{2} = q + 0.75\xi $$Final Results
Recall that \(\xi = \frac{rq}{a} \left( \frac{2-\sqrt{2}}{4} \right)\). Substituting this back into our results:
Final Charge on B (Corner Sphere):
$$ Q_B = q – 1.5\xi = q \left\{ 1 – \frac{3}{2} \cdot \frac{r}{a} \left( \frac{2-\sqrt{2}}{4} \right) \right\} $$ $$ \mathbf{Q_B \approx q \left\{ 1 – \frac{3r}{2a} \left( \frac{2-\sqrt{2}}{4} \right) \right\}} $$(Matches the formula for “Final Charge on A” in the reference solution, confirming A was the corner in that version)
Final Charge on A and C (Acute Spheres):
$$ Q_A = Q_C = q + 0.75\xi = q \left\{ 1 + \frac{3}{4} \cdot \frac{r}{a} \left( \frac{2-\sqrt{2}}{4} \right) \right\} $$ $$ \mathbf{Q_A = Q_C \approx q \left\{ 1 + \frac{3r}{4a} \left( \frac{2-\sqrt{2}}{4} \right) \right\}} $$(Matches the formula for “Final Charge on B and C” in the reference solution)