Problem Statement
Determine the electric field for the configuration described: A particle is given a charge of $0.2\times10^{-6}$ C and placed in an electric field of $3000$ N/C. Find the electric force on the particle. $F = qE$ Step 1: $q = 2.0\times10^{-7}$ C, $E = 3000$ N/C. Step 2: $$F = qE = 2.0\times10^{-7} \times 3000 = 6.0\times10^{-4} \text{ N}$$ $$\boxed{F = 6.0\time
Given Information
- See problem statement for all given quantities.
Physical Concepts & Formulas
Gauss’s law relates the electric flux through any closed surface to the total enclosed charge. It is one of Maxwell’s four equations and is especially powerful when the charge distribution has spherical, cylindrical, or planar symmetry, because the flux integral then simplifies to $E \cdot A = Q_\text{enc}/\varepsilon_0$.
- $\oint \mathbf{E}\cdot d\mathbf{A} = Q_{\text{enc}}/\varepsilon_0$ — Gauss’s law
- $E = Q/(4\pi\varepsilon_0 r^2)$ — field outside a sphere
- $E = \sigma/\varepsilon_0$ — field between infinite parallel plates
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$F = qE = 2.0\times10^{-7} \times 3000 = 6.0\times10^{-4} \text{ N}$$
$$\boxed{F = 6.0\time
Given Information
- Geometry and charge distribution as given in the problem
- Permittivity of free space $\varepsilon_0 = 8.85\times10^{-12}\,\text{C}^2\text{N}^{-1}\text{m}^{-2}$
- Coulomb constant $k = 1/(4\pi\varepsilon_0) = 9\times10^9\,\text{N m}^2\text{C}^{-2}$
Physical Concepts & Formulas
The electric field $\vec{E}$ at any point is the force per unit positive test charge placed at that point. For symmetric charge distributions, Gauss’s Law — $\oint \vec{E}\cdot d\vec{A} = Q_{\text{enc}}/\varepsilon_0$ — is the most powerful tool: choose a Gaussian surface matching the symmetry, and the flux integral reduces to $E \times A$. For arbitrary distributions, direct integration using Coulomb’s law with superposition gives $\vec{E} = \int k\,dq\,\hat{r}/r^2$. The direction of $\vec{E}$ always points away from positive charges and toward negative charges.
- $\vec{E} = \dfrac{kq}{r^2}\hat{r}$ — Coulomb’s law for a point charge
- $\oint \vec{E}\cdot d\vec{A} = \dfrac{Q_{\text{enc}}}{\varepsilon_0}$ — Gauss’s Law
- For a uniformly charged sphere (outside): $E = \dfrac{kQ}{r^2}$
- For an infinite line charge: $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$
- For an infinite plane: $E = \dfrac{\sigma}{2\varepsilon_0}$
Step-by-Step Solution
Step 1 — Identify symmetry: Spherical, cylindrical, or planar symmetry determines which Gaussian surface to use.
Step 2 — Choose Gaussian surface: Select a surface where $\vec{E}$ is either parallel or perpendicular to $d\vec{A}$ everywhere, and where $|E|$ is constant on the surface.
Step 3 — Apply Gauss’s Law:
$$
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Step 4 — Solve for $E$:
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Answer
$$\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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