Electric Fields Of Charge Distributions

See the full derivation above. Starting from dE = (1/4πε₀)(dq/s²), express dq = λ dy and s = √(r² + y²). State that by symmetry the components parallel to the wire cancel, leaving only the perpendicular component. Set up the integral of dE_perp from −∞ to +∞. Evaluate using the standard integral ∫dy/(r² + y²)^(3/2) = 2/r². Simplify to obtain E = λ/(2πε₀r).

The field decreases. Because E is inversely proportional to r, tripling r reduces E to one-third of its original value.

Begin with dE = (1/4πε₀)(dq/s²). Set dq = (Q/L)dy and s = √(d² + y²). By symmetry, the y-components cancel. The surviving x-component is dE_x = (Q d dy)/(4πε₀ L (d² + y²)^(3/2)). Integrate from −L/2 to +L/2. Evaluate using the standard integral to obtain E = Q/(4πε₀ d √(d² + L²/4)), directed perpendicular to the rod and away from it.

The expression is consistent. When d ≫ L, the term L²/4 under the square root becomes negligible compared to d². The denominator becomes d · d = d², so E → Q/(4πε₀d²). This matches Coulomb’s law for a point charge, which is expected because at large distances the finite rod appears as a point.

Equation used $$E=\frac{Q}{4\pi {\epsilon }_{0} d\sqrt{{d}^{2}+{L}^{2}/4}}$$ Given $$Q=4.0\times {10}^{-9}\text{ C}$$ $$d=0.40\text{ m}$$ $$L=0.30\text{ m}$$ Finding the field $${d}^{2}+\frac{{L}^{2}}{4}=0.16+0.0225=0.1825$$ $$\sqrt{0.1825}=0.4272\text{ m}$$ $$E=\frac{4.0\times {10}^{-9}}{4\pi (8.85\times {10}^{-12})(0.40)(0.4272)}$$ $$E=\frac{4.0\times {10}^{-9}}{1.902\times {10}^{-11}}$$ $$E\approx 210\text{ N/C}$$ The field is approximately 210 N/C, directed perpendicular to the rod and away from it. Interpretation: The point-charge value Q/(4πε₀d²) ≈ 225 N/C. The finite-rod result (210 N/C) is slightly less because the off-center elements contribute at an angle, reducing the net perpendicular component.

Starting point: dE = (1/4πε₀)(dq/s²) and superposition. Every element is at distance s = √(x² + R²) from the field point. By symmetry, radial components cancel. The axial component is dE_x = (dq)(x)/(4πε₀(x² + R²)^(3/2)). Since s is constant for all elements, integrate: E = x/(4πε₀(x² + R²)^(3/2)) × ∫dq = Qx/(4πε₀(x² + R²)^(3/2)).

Differentiate E with respect to x and set dE/dx = 0. Using the quotient rule on Qx(x² + R²)^(−3/2): $$\frac{dE}{dx}∝({x}^{2}+{R}^{2}{)}^{-3/2}-3{x}^{2}({x}^{2}+{R}^{2}{)}^{-5/2}=0$$ $$({x}^{2}+{R}^{2})-3{x}^{2}=0$$ $${R}^{2}=2{x}^{2}$$ $$x=\frac{R}{\sqrt{2}}$$ The maximum field occurs at x = R/√2. Beyond this point, the 1/x² falloff dominates. Closer to the ring, the angle reduces the axial component.

Starting point: dE = (1/4πε₀)(dq/R²) and superposition. Set dq = λR dθ. Every element is at distance R from the center. By symmetry, components perpendicular to the axis of symmetry cancel. The surviving component is dE_x = (λ cos θ dθ)/(4πε₀R). Integrate from −π/2 to +π/2: ∫cos θ dθ = 2. Result: E = λ/(2πε₀R), directed along the symmetry axis.

The symmetry changes. Both x- and y-components survive because the arc no longer has mirror symmetry about any single axis through the center. Integrating dE_x = (λ cos θ dθ)/(4πε₀R) from 0 to π/2 gives ∫cos θ dθ = 1, so $E_{\mathrm{x}}$ = λ/(4πε₀R). Integrating dE_y = (λ sin θ dθ)/(4πε₀R) from 0 to π/2 gives ∫sin θ dθ = 1, so $E_{\mathrm{y}}$ = λ/(4πε₀R). The magnitude is E = λ√2/(4πε₀R), and the direction is at 45° from the x-axis — bisecting the arc, unlike the semicircular case where the field lies along one axis only.

Starting point: Gauss’s law ∮E⃗·dA⃗ = $Q_{\mathrm{enc}}$/ε₀. Choose a coaxial cylindrical Gaussian surface of radius r < a and length L. By cylindrical symmetry, E is constant on the curved surface and perpendicular to it, so the flux is E(2πrL). The end caps contribute zero flux because E is parallel to them. The enclosed charge is $Q_{\mathrm{enc}}$ = ρπr²L. Setting the flux equal to $Q_{\mathrm{enc}}$/ε₀: E(2πrL) = ρπr²L/ε₀. Solving: E = ρr/(2ε₀), directed radially outward.

The field at r = a/2 equals the field at r = 2a. Inside the cylinder at r = a/2: $$E=\frac{\rho (a/2)}{2{\epsilon }_{0}}=\frac{\rho a}{4{\epsilon }_{0}}$$ Outside the cylinder at r = 2a: $$E=\frac{\rho {a}^{2}}{2{\epsilon }_{0}(2a)}=\frac{\rho a}{4{\epsilon }_{0}}$$ The two values are equal. This is because the linear increase inside and the 1/r decrease outside happen to give the same value at these symmetric points.

Inside the shell (r < a), $Q_{\mathrm{enc}}$ = 0, so E = 0 everywhere. The linear region disappears entirely. At r = a, the field jumps discontinuously from 0 to λ/(2πε₀a). For r > a, the field is the same as before: E = λ/(2πε₀r), decreasing as 1/r. The graph would show a flat line at E = 0 for r < a, a vertical jump at r = a, and the same 1/r curve for r > a.