Problem Statement
Solve the capacitor/capacitance problem: A parallel plate capacitor with plate area $A = 0.02$ m$^2$ and separation $d = 2$ mm is connected to a 12 V battery. Find the charge on each plate. ($\varepsilon_0 = 8.85\times10^{-12}$ F/m) Capacitance: $C = \varepsilon_0 A/d$ Charge: $Q = CV$ Step 1: $$C = \frac{\varepsilon_0 A}{d} = \frac{8.85\t
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Capacitors store electric charge on conducting plates separated by an insulator (dielectric). The capacitance $C = Q/V$ depends on geometry and dielectric constant. Energy stored is $U = Q^2/(2C) = CV^2/2 = QV/2$. Series and parallel combinations follow rules opposite to resistors.
- $C = Q/V$ — definition of capacitance
- $U = \frac{1}{2}CV^2 = \frac{Q^2}{2C}$ — energy stored
- $C_{\text{parallel}} = C_1 + C_2$ — parallel combination
- $1/C_{\text{series}} = 1/C_1 + 1/C_2$ — series combination
- $C = \varepsilon_0\varepsilon_r A/d$ — parallel plate capacitor
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$C = \frac{\varepsilon_0 A}{d} = \frac{8.85\t
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: $A = 0.02\,\text{m}^2$, $d = 1\,\text{mm} = 10^{-3}\,\text{m}$, $\kappa = 2$:
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Answer
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Answer
$$\boxed{C = \dfrac{\kappa\varepsilon_0 A}{d}}$$
Physical Interpretation
Capacitors store energy in the electric field between their plates. Doubling the voltage quadruples the stored energy — an important design constraint for high-voltage applications. Charge sharing between capacitors is a lossless process only in the ideal case; real circuits dissipate energy in connecting resistance.
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