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Gauss's Law
Gauss's Law and its Application to a Sphere's Field
Gauss's Law
Gauss's Law in electromagnetism relates the electric flux through a closed surface to the charge enclosed within that surface. The mathematical formula is:
\[ Φ_E = ∮E· dA =\frac { Q_{int}}{ ε_₀}\]
Where:
\[ Φ_E\]
is the electric flux.
\[E\]
is the electric field.
\[Q_{int}\]
is the enclosed charge.
\[ε_₀\]
is the vacuum permittivity.
Calculating the Electric Field of a Uniformly Charged Sphere
1. Inside the sphere
\[(R >r)\]
Enclosed charge:
\[ Q_{int} = \frac {(Q × r³)}{ R³}\]
Electric field:
\[ E = \frac {(k × Q × r)}{ R³}\]
Where
\[k = 1/(4πε_₀)\]
2. On the sphere's surface
\[(r = R)\]
Electric field:
\[E = \frac {(k × Q) }{ R²}\]
3. Outside the sphere
\[(r > R)\]
Electric field:
\[ E =\frac {(k × Q) }{ r²}\]
Practical Applications
- Design of spherical capacitors in electronic circuits
- Calculating electric fields of stars and planets (in astrophysics)
- Electrical insulation systems in laboratories
- Van de Graaff generator (uses a metal sphere to store charges)
Calculation Example
If the sphere's charge is
\[Q = 2 × 10^{-6}\; C \]and its radius is \[R = 0.5\; m\]
- Field at \[r = 0.3 \;m\]
\[E = 8.99 × 10^9 ×\frac { 2 × 10^{-6} × 0.3 }{ (0.5)^3}≈ 43152\; N/C\]
- Field on the surface
\[ E = 8.99 × 10^9 × \frac {2 × 10^{-6}}{ (0.5)^2} ≈ 71920\; N/C\]
- Field at
\[r = 1 m\]\[ E = 8.99 × 10^9×\frac {2 × 10^{-6}} {(1)^2} = 17980\; N/C\]
Calculating the field inside the sphere\[E=\frac{K.q_t.r_1}{R^3}\]
Calculating the field outside the sphere \[E=\frac{k.q_t}{r^2}\]
Calculating the Electric Field of a Charged Sphere
Electric Field Calculator for a Charged Sphere
Electric Field of a Charged Wire
Electric Field of a Finite Length Charged Wire
Basic Equations
Linear charge density
\[(λ) = \frac {Q}{L}\]
Electric field at point
\[P\]
at distance
\[ r\]
from the wire:
\[E = \frac {1}{4πε₀} * \frac {λ}{r} [sinθ₁ + sinθ₂]\]
Mathematical Derivation
1. Divide the wire into small elements
\[(dx)\]
2. Calculate field from each element using Coulomb's Law
3. Integrate along the wire using:
\[∫ dE = (\frac {λ}{(4πε₀)}) .∫ (dx)/(r² + x²)^{3/2}\]
Practical Applications
- Design of safe electrical wiring
- Electromagnetic field protection systems
- Micro electrical sensors
- Applications in medical physics (e.g., ECG devices)
Calculation Example
A 2m long wire carries 4μC charge:
λ = 4μC / 2m = 2μC/m
If θ₁ = θ₂ = 45°, r = 50cm
\[E = (9×10^9) ×\frac {2×10^{-6}}{0.5}× [sin45 + sin45] ≈ 50.9 \;kN/C\]
Important note:
When L → ∞ (infinite wire):
\[E = \frac {(λ)}{(2πε₀r)}\]
Gauss's Law |
Gauss's Law
Gauss's Law in electromagnetism relates the electric flux through a closed surface to the charge enclosed within that surface. The mathematical formula is:
\[ Φ_E = ∮E· dA =\frac { Q_{int}}{ ε_₀}\]
Where:
\[ Φ_E\]
is the electric flux.
\[E\]
is the electric field.
\[Q_{int}\]
is the enclosed charge.
\[ε_₀\]
is the vacuum permittivity.
Calculating the Electric Field of a Uniformly Charged Sphere
1. Inside the sphere \[(R >r)\] Enclosed charge: \[ Q_{int} = \frac {(Q × r³)}{ R³}\] Electric field: \[ E = \frac {(k × Q × r)}{ R³}\] Where \[k = 1/(4πε_₀)\]2. On the sphere's surface \[(r = R)\]
Electric field: \[E = \frac {(k × Q) }{ R²}\]
3. Outside the sphere \[(r > R)\]
Electric field: \[ E =\frac {(k × Q) }{ r²}\]
Practical Applications
- Design of spherical capacitors in electronic circuits
- Calculating electric fields of stars and planets (in astrophysics)
- Electrical insulation systems in laboratories
- Van de Graaff generator (uses a metal sphere to store charges)
Calculation Example
If the sphere's charge is
\[Q = 2 × 10^{-6}\; C \]and its radius is \[R = 0.5\; m\]
- Field at \[r = 0.3 \;m\]
\[E = 8.99 × 10^9 ×\frac { 2 × 10^{-6} × 0.3 }{ (0.5)^3}≈ 43152\; N/C\]
- Field on the surface
\[ E = 8.99 × 10^9 × \frac {2 × 10^{-6}}{ (0.5)^2} ≈ 71920\; N/C\]
- Field at
\[r = 1 m\]\[ E = 8.99 × 10^9×\frac {2 × 10^{-6}} {(1)^2} = 17980\; N/C\]
Electric Field Calculator for a Charged Sphere
Electric Field of a Finite Length Charged Wire
Basic Equations
Linear charge density \[(λ) = \frac {Q}{L}\]
Electric field at point \[P\] at distance \[ r\] from the wire:
\[E = \frac {1}{4πε₀} * \frac {λ}{r} [sinθ₁ + sinθ₂]\]
Mathematical Derivation
1. Divide the wire into small elements \[(dx)\]
2. Calculate field from each element using Coulomb's Law
3. Integrate along the wire using:
\[∫ dE = (\frac {λ}{(4πε₀)}) .∫ (dx)/(r² + x²)^{3/2}\]
Practical Applications
- Design of safe electrical wiring
- Electromagnetic field protection systems
- Micro electrical sensors
- Applications in medical physics (e.g., ECG devices)
Calculation Example
A 2m long wire carries 4μC charge:
λ = 4μC / 2m = 2μC/m
If θ₁ = θ₂ = 45°, r = 50cm
\[E = (9×10^9) ×\frac {2×10^{-6}}{0.5}× [sin45 + sin45] ≈ 50.9 \;kN/C\]
Important note:
When L → ∞ (infinite wire):
\[E = \frac {(λ)}{(2πε₀r)}\]
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