Problem Statement
Determine the electric field for the configuration described: Electrons with speed $v = 3\times10^7$ m/s enter a uniform electric field $E = 10^4$ V/m perpendicular to their velocity. Find the transverse deflection after travelling $L = 0.1$ m horizontally. Time to cross: $t = L/v$; transverse acceleration: $a = eE/m_e$; deflection: $y = \frac{1}{2}at^2$ Step
Given Information
- See problem statement for all given quantities.
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
$$E \cdot A_{\text{surface}} = \frac{Q_{\text{enc}}}{\varepsilon_0}$$
$$E = \frac{Q_{\text{enc}}}{\varepsilon_0 A_{\text{surface}}}$$
$$E(4\pi r^2) = \frac{Q}{\varepsilon_0} \implies E = \frac{Q}{4\pi\varepsilon_0 r^2} = \frac{kQ}{r^2}$$
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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