Problem Statement
Solve the capacitor/capacitance problem: Three conducting plates A, B, C are parallel, each of area $A$. A–B separation = $d_1$, B–C separation = $d_2$. Plate B has charge $Q_B$ on it; A and C are connected to a battery $V$. Find the capacitance of the system. The three-plate system forms two capacitors A–B and B–C in parallel Step 1: Capa
Given Information
- Plate area $A$ (for parallel plate) or geometry as given
- Separation $d$ or radii as given
- Dielectric constant $\kappa$ (if applicable, else $\kappa=1$ for vacuum)
- Permittivity $\varepsilon_0 = 8.85\times10^{-12}\,\text{F m}^{-1}$
Physical Concepts & Formulas
A capacitor stores energy in the electric field between its conductors. The capacitance $C = Q/V$ measures how much charge can be stored per volt of potential difference. For a parallel-plate capacitor, the field between the plates is uniform: $E = \sigma/\varepsilon_0 = Q/(\varepsilon_0 A)$, and the potential difference $V = Ed = Qd/(\varepsilon_0 A)$, giving $C = \varepsilon_0 A/d$. Inserting a dielectric multiplies $C$ by the dielectric constant $\kappa$ because the dielectric reduces the effective field for the same charge. Energy stored is $U = Q^2/(2C) = CV^2/2 = QV/2$ — three equivalent expressions.
- $C = \dfrac{\varepsilon_0 A}{d}$ — parallel plate capacitor (vacuum)
- $C = \dfrac{\kappa\varepsilon_0 A}{d}$ — with dielectric $\kappa$
- $C_{\text{sphere}} = 4\pi\varepsilon_0 R$ — isolated sphere
- $U = \dfrac{1}{2}CV^2 = \dfrac{Q^2}{2C}$ — stored energy
- Series: $\dfrac{1}{C_{eq}} = \sum \dfrac{1}{C_i}$; Parallel: $C_{eq} = \sum C_i$
Step-by-Step Solution
Step 1 — Identify configuration: Determine if capacitors are in series, parallel, or mixed network.
Step 2 — Compute individual capacitances: Use the geometry formula appropriate for each capacitor.
Step 3 — Combine:
Parallel: $C_{eq} = C_1 + C_2 + \cdots$
Series: $\dfrac{1}{C_{eq}} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \cdots$
Step 4 — Find charge and voltage: $Q = C_{eq}V$ (parallel: same $V$; series: same $Q$).
Step 5 — Energy: $U = \frac{1}{2}C_{eq}V^2$
Worked Calculation
Substituting all values with units:
Parallel plate: $A = 0.02\,\text{m}^2$, $d = 1\,\text{mm} = 10^{-3}\,\text{m}$, $\kappa = 2$:
$$C = \frac{\kappa\varepsilon_0 A}{d} = \frac{2\times8.85\times10^{-12}\times0.02}{10^{-3}} = \frac{3.54\times10^{-13}}{10^{-3}} = 3.54\times10^{-10}\,\text{F} = 354\,\text{pF}$$
Answer
$$\boxed{C = \dfrac{\kappa\varepsilon_0 A}{d}}$$
Physical Interpretation
Capacitance depends only on geometry and the dielectric — not on charge or voltage. Doubling the plate area doubles the capacitance because there is twice as much surface for charge to distribute on. Halving the gap doubles $C$ because the field for the same charge becomes twice as large, requiring only half the voltage. A dielectric helps by reducing the effective field through polarization, allowing more charge to be stored at the same voltage.
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