Problem Statement
Solve the capacitor/capacitance problem: Two batteries ($V_1 = 6$ V, $V_2 = 4$ V) and a capacitor $C = 2\mu$F are in a loop (batteries opposing, capacitor between them). Find the charge on the capacitor. See problem statement for all given quantities. Capacitors store electric charge on conducting plates separated by an insulator (dielectr
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.
Leave a Reply