Problem Statement
Two conducting spheres of radii $R_1$ and $R_2$ are connected by a thin wire far from each other. A total charge $Q$ is placed on the system. Find the charge on each sphere and the electric field at each sphere’s surface.
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
Two conducting spheres of radii $R_1$ and $R_2$ are connected by a thin wire far from each other. A total charge $Q$ is placed on the system. Find the charge on each sphere and the electric field at each sphere’s surface.
Given Information
- $R_1$, $R_2$ = sphere radii
- $Q = Q_1 + Q_2$ = total charge
- Spheres connected → same potential
Physical Concepts & Formulas
When conductors are connected, they reach the same potential. The potential of an isolated sphere is $kQ/R$ (since charge sits uniformly on a conductor surface). Equating potentials and using charge conservation gives two equations for the two unknowns $Q_1$, $Q_2$.
- $\varphi_1 = \varphi_2$ — equipotential condition
- $\varphi = kQ_i/R_i$ — potential of charged sphere
Step-by-Step Solution
Step 1 — Equipotential condition:
$$\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \implies \frac{Q_1}{R_1} = \frac{Q_2}{R_2}$$
Step 2 — Charge conservation: $Q_1 + Q_2 = Q$
Step 3 — Solve: From equipotential: $Q_1 = Q_2 R_1/R_2$. Substituting:
$$Q_2\frac{R_1}{R_2} + Q_2 = Q \implies Q_2 = \frac{QR_2}{R_1+R_2}, \quad Q_1 = \frac{QR_1}{R_1+R_2}$$
Step 4 — Surface fields:
$$E_1 = \frac{kQ_1}{R_1^2} = \frac{kQ}{R_1(R_1+R_2)}, \quad E_2 = \frac{kQ_2}{R_2^2} = \frac{kQ}{R_2(R_1+R_2)}$$
$$\frac{E_1}{E_2} = \frac{R_2}{R_1}$$
Worked Calculation
For $R_1 = 1\,\text{cm}$, $R_2 = 4\,\text{cm}$, $Q = 10\,\mu\text{C}$:
$$Q_1 = \frac{10 \times 1}{1+4} = 2\,\mu\text{C}, \quad Q_2 = 8\,\mu\text{C}$$
$$E_1/E_2 = 4 \text{ (smaller sphere has stronger surface field)}$$
Answer
$$\boxed{Q_1 = \frac{QR_1}{R_1+R_2}, \quad Q_2 = \frac{QR_2}{R_1+R_2}, \quad \frac{E_1}{E_2} = \frac{R_2}{R_1}}$$
Physical Interpretation
Smaller spheres accumulate stronger surface fields even though they hold less charge — this is why sharp points on conductors concentrate field (and can cause corona discharge). Lightning rods are pointed to maximize the local field and initiate controlled discharge. A sphere of radius 1 mm at the same potential as one of radius 1 m has a 1000× stronger surface field.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\boxed{Q_1 = \frac{QR_1}{R_1+R_2}, \quad Q_2 = \frac{QR_2}{R_1+R_2}, \quad \frac{E_1}{E_2} = \frac{R_2}{R_1}}}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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