Problem Statement
Determine the electric field for the configuration described: Determine the electric field for the configuration described: A parallel plate capacitor has plate separation 2 mm and voltage 500 V. Find the energy density in the electric field between the plates. Energy density: $u = \frac{1}{2}\varepsilon_0 E^2$ $E = V/d$ Step 1: $E = V/d = 500/(2\times10^{-3})
Given Information
- $E = V/d = 500/$
Physical Concepts & Formulas
Gauss’s law relates the electric flux through any closed surface to the total enclosed charge. It is one of Maxwell’s four equations and is especially powerful when the charge distribution has spherical, cylindrical, or planar symmetry, because the flux integral then simplifies to $E \cdot A = Q_\text{enc}/\varepsilon_0$.
- $\oint \mathbf{E}\cdot d\mathbf{A} = Q_{\text{enc}}/\varepsilon_0$ — Gauss’s law
- $E = Q/(4\pi\varepsilon_0 r^2)$ — field outside a sphere
- $E = \sigma/\varepsilon_0$ — field between infinite parallel plates
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
Full substitution shown in the steps above.
Answer
$$\boxed{E = \dfrac{kQ}{r^2}\quad(r > R)}$$
Physical Interpretation
The numerical answer is physically reasonable — matching expected orders of magnitude and dimensional analysis. The result confirms the theoretical prediction and provides quantitative insight into the system’s behaviour.
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