Gauss's Law

Integral form of Gauss's law and applications to symmetric charge distributions

Gauss's Law

Statement of Gauss's Law

∮E⃗⋅dA⃗=Qencϵ0\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}

The electric flux through a closed surface equals enclosed charge divided by ϵ0\epsilon_0.

Electric Flux

ΦE=∫E⃗⋅dA⃗\Phi_E = \int \vec{E} \cdot d\vec{A}

For uniform field and flat surface: ΦE=E⃗⋅A⃗=EAcos⁡θ\Phi_E = \vec{E} \cdot \vec{A} = EA\cos\theta

Applying Gauss's Law

Strategy:

  1. Choose Gaussian surface with symmetry
  2. Calculate flux on each part
  3. Find enclosed charge
  4. Solve for EE

Works best for:

  • Spherical symmetry
  • Cylindrical symmetry
  • Planar symmetry

Spherical Symmetry

Point Charge

Gaussian surface: sphere of radius rr

∮E dA=E(4πr2)=qϵ0\oint E \, dA = E(4\pi r^2) = \frac{q}{\epsilon_0}

E=14πϵ0qr2=kqr2E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} = k\frac{q}{r^2}

Uniform Spherical Shell

Shell of radius RR, total charge QQ:

Outside (r>Rr > R): E(4πr2)=Qϵ0E(4\pi r^2) = \frac{Q}{\epsilon_0}

E=kQr2E = k\frac{Q}{r^2}

Inside (r<Rr < R): Qenc=0⇒E=0Q_{enc} = 0 \Rightarrow E = 0

Uniform Solid Sphere

Sphere of radius RR, uniform charge density ρ\rho, total charge QQ:

Outside (r>Rr > R): E=kQr2E = k\frac{Q}{r^2}

Inside (r<Rr < R): Qenc=ρ⋅43πr3=Qr3R3Q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = Q\frac{r^3}{R^3}

E(4πr2)=Qϵ0r3R3E(4\pi r^2) = \frac{Q}{\epsilon_0}\frac{r^3}{R^3}

E=kQR3rE = k\frac{Q}{R^3}r

(Linear in rr, maximum at surface)

Cylindrical Symmetry

Infinite Line Charge

Line with charge density λ\lambda:

Gaussian surface: cylinder of radius rr, length LL

E(2πrL)=λLϵ0E(2\pi rL) = \frac{\lambda L}{\epsilon_0}

E=λ2πϵ0r=2kλrE = \frac{\lambda}{2\pi\epsilon_0 r} = \frac{2k\lambda}{r}

Infinite Cylindrical Shell

Shell of radius RR, charge per length λ\lambda:

Outside (r>Rr > R): E=λ/(2πϵ0r)E = \lambda/(2\pi\epsilon_0 r)

Inside (r<Rr < R): E=0E = 0

Infinite Solid Cylinder

Cylinder of radius RR, uniform volume charge density ρ\rho:

Outside (r>Rr > R): E=λ/(2πϵ0r)E = \lambda/(2\pi\epsilon_0 r) where λ=ρπR2\lambda = \rho\pi R^2

Inside (r<Rr < R): Qenc=ρ(πr2L)Q_{enc} = \rho(\pi r^2 L)

E=ρr2ϵ0E = \frac{\rho r}{2\epsilon_0}

Planar Symmetry

Infinite Sheet

Surface charge density σ\sigma:

Gaussian surface: pillbox with area AA

2EA=σAϵ02EA = \frac{\sigma A}{\epsilon_0}

E=σ2ϵ0E = \frac{\sigma}{2\epsilon_0}

(Field is uniform, independent of distance!)

Two Parallel Sheets

Sheets with +σ+\sigma and −σ-\sigma:

Between sheets: E=σ/ϵ0E = \sigma/\epsilon_0 (fields add)

Outside: E=0E = 0 (fields cancel)

This is a capacitor.

Conductors in Electrostatic Equilibrium

  1. E⃗=0\vec{E} = 0 inside conductor
  2. Net charge resides on surface
  3. E⃗\vec{E} perpendicular to surface just outside
  4. E=σ/ϵ0E = \sigma/\epsilon_0 just outside surface

Conducting Shell

Hollow conductor with charge QQ:

  • Inner surface: charge −q-q (if +q+q inside cavity)
  • Outer surface: charge Q+qQ + q
  • Field inside conductor: E=0E = 0
  • Field in cavity: depends on charge inside

Differential Form

Using divergence theorem:

∇⋅E⃗=ρϵ0\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}

This is one of Maxwell's equations.

📚 Practice Problems

1Problem 1medium

❓ Question:

A solid insulating sphere of radius R = 0.08 m has a uniform volume charge density ρ = 6.5 × 10⁻⁶ C/m³. Use Gauss's law to find: (a) the electric field at r = 0.05 m (inside), (b) the electric field at r = 0.12 m (outside), and (c) the total charge of the sphere.

💡 Show Solution

Given:

  • R = 0.08 m
  • ρ = 6.5 × 10⁻⁶ C/m³
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
  • k = 8.99 × 10⁹ N·m²/C²

Gauss's Law: ∮E⃗·dA⃗ = Q_enc/ε₀

(a) Inside sphere (r = 0.05 m < R):

Choose Gaussian surface: sphere of radius r

By symmetry, E is constant on surface and radial: E(4πr2)=Qencε0E(4πr^2) = \frac{Q_{enc}}{ε_0}

Enclosed charge: Qenc=ρ⋅43πr3Q_{enc} = ρ \cdot \frac{4}{3}πr^3

E=ρr33ε0r2=ρr3ε0E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}

E=(6.5×10−6)(0.05)3(8.85×10−12)=3.25×10−72.655×10−11E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

Or using k = 1/(4πε₀): E=4πkρr3=4π(8.99×109)(6.5×10−6)(0.05)3E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: Q=ρ⋅43πR3=(6.5×10−6)⋅43π(0.08)3Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3

Q=(6.5×10−6)(2.144×10−3)=1.39×10−8 CQ = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}

Outside, field is like a point charge: E=kQr2=(8.99×109)(1.39×10−8)(0.12)2E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}

E=1.25×1021.44×10−2=8.68×103 N/CE = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

  • From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
  • From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC

2Problem 2medium

❓ Question:

A solid insulating sphere of radius R = 0.08 m has a uniform volume charge density ρ = 6.5 × 10⁻⁶ C/m³. Use Gauss's law to find: (a) the electric field at r = 0.05 m (inside), (b) the electric field at r = 0.12 m (outside), and (c) the total charge of the sphere.

💡 Show Solution

Given:

  • R = 0.08 m
  • ρ = 6.5 × 10⁻⁶ C/m³
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
  • k = 8.99 × 10⁹ N·m²/C²

Gauss's Law: ∮E⃗·dA⃗ = Q_enc/ε₀

(a) Inside sphere (r = 0.05 m < R):

Choose Gaussian surface: sphere of radius r

By symmetry, E is constant on surface and radial: E(4πr2)=Qencε0E(4πr^2) = \frac{Q_{enc}}{ε_0}

Enclosed charge: Qenc=ρ⋅43πr3Q_{enc} = ρ \cdot \frac{4}{3}πr^3

E=ρr33ε0r2=ρr3ε0E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}

E=(6.5×10−6)(0.05)3(8.85×10−12)=3.25×10−72.655×10−11E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

Or using k = 1/(4πε₀): E=4πkρr3=4π(8.99×109)(6.5×10−6)(0.05)3E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: Q=ρ⋅43πR3=(6.5×10−6)⋅43π(0.08)3Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3

Q=(6.5×10−6)(2.144×10−3)=1.39×10−8 CQ = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}

Outside, field is like a point charge: E=kQr2=(8.99×109)(1.39×10−8)(0.12)2E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}

E=1.25×1021.44×10−2=8.68×103 N/CE = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

  • From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
  • From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC

3Problem 3medium

❓ Question:

A solid insulating sphere of radius R = 0.08 m has a uniform volume charge density ρ = 6.5 × 10⁻⁶ C/m³. Use Gauss's law to find: (a) the electric field at r = 0.05 m (inside), (b) the electric field at r = 0.12 m (outside), and (c) the total charge of the sphere.

💡 Show Solution

Given:

  • R = 0.08 m
  • ρ = 6.5 × 10⁻⁶ C/m³
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
  • k = 8.99 × 10⁹ N·m²/C²

Gauss's Law: ∮E⃗·dA⃗ = Q_enc/ε₀

(a) Inside sphere (r = 0.05 m < R):

Choose Gaussian surface: sphere of radius r

By symmetry, E is constant on surface and radial: E(4πr2)=Qencε0E(4πr^2) = \frac{Q_{enc}}{ε_0}

Enclosed charge: Qenc=ρ⋅43πr3Q_{enc} = ρ \cdot \frac{4}{3}πr^3

E=ρr33ε0r2=ρr3ε0E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}

E=(6.5×10−6)(0.05)3(8.85×10−12)=3.25×10−72.655×10−11E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

Or using k = 1/(4πε₀): E=4πkρr3=4π(8.99×109)(6.5×10−6)(0.05)3E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: Q=ρ⋅43πR3=(6.5×10−6)⋅43π(0.08)3Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3

Q=(6.5×10−6)(2.144×10−3)=1.39×10−8 CQ = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}

Outside, field is like a point charge: E=kQr2=(8.99×109)(1.39×10−8)(0.12)2E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}

E=1.25×1021.44×10−2=8.68×103 N/CE = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

  • From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
  • From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC

4Problem 4medium

❓ Question:

A solid insulating sphere of radius R = 0.08 m has a uniform volume charge density ρ = 6.5 × 10⁻⁶ C/m³. Use Gauss's law to find: (a) the electric field at r = 0.05 m (inside), (b) the electric field at r = 0.12 m (outside), and (c) the total charge of the sphere.

💡 Show Solution

Given:

  • R = 0.08 m
  • ρ = 6.5 × 10⁻⁶ C/m³
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
  • k = 8.99 × 10⁹ N·m²/C²

Gauss's Law: ∮E⃗·dA⃗ = Q_enc/ε₀

(a) Inside sphere (r = 0.05 m < R):

Choose Gaussian surface: sphere of radius r

By symmetry, E is constant on surface and radial: E(4πr2)=Qencε0E(4πr^2) = \frac{Q_{enc}}{ε_0}

Enclosed charge: Qenc=ρ⋅43πr3Q_{enc} = ρ \cdot \frac{4}{3}πr^3

E=ρr33ε0r2=ρr3ε0E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}

E=(6.5×10−6)(0.05)3(8.85×10−12)=3.25×10−72.655×10−11E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

Or using k = 1/(4πε₀): E=4πkρr3=4π(8.99×109)(6.5×10−6)(0.05)3E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: Q=ρ⋅43πR3=(6.5×10−6)⋅43π(0.08)3Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3

Q=(6.5×10−6)(2.144×10−3)=1.39×10−8 CQ = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}

Outside, field is like a point charge: E=kQr2=(8.99×109)(1.39×10−8)(0.12)2E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}

E=1.25×1021.44×10−2=8.68×103 N/CE = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

  • From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
  • From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC

5Problem 5medium

❓ Question:

A solid insulating sphere of radius R = 0.08 m has a uniform volume charge density ρ = 6.5 × 10⁻⁶ C/m³. Use Gauss's law to find: (a) the electric field at r = 0.05 m (inside), (b) the electric field at r = 0.12 m (outside), and (c) the total charge of the sphere.

💡 Show Solution

Given:

  • R = 0.08 m
  • ρ = 6.5 × 10⁻⁶ C/m³
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
  • k = 8.99 × 10⁹ N·m²/C²

Gauss's Law: ∮E⃗·dA⃗ = Q_enc/ε₀

(a) Inside sphere (r = 0.05 m < R):

Choose Gaussian surface: sphere of radius r

By symmetry, E is constant on surface and radial: E(4πr2)=Qencε0E(4πr^2) = \frac{Q_{enc}}{ε_0}

Enclosed charge: Qenc=ρ⋅43πr3Q_{enc} = ρ \cdot \frac{4}{3}πr^3

E=ρr33ε0r2=ρr3ε0E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}

E=(6.5×10−6)(0.05)3(8.85×10−12)=3.25×10−72.655×10−11E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

Or using k = 1/(4πε₀): E=4πkρr3=4π(8.99×109)(6.5×10−6)(0.05)3E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: Q=ρ⋅43πR3=(6.5×10−6)⋅43π(0.08)3Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3

Q=(6.5×10−6)(2.144×10−3)=1.39×10−8 CQ = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}

Outside, field is like a point charge: E=kQr2=(8.99×109)(1.39×10−8)(0.12)2E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}

E=1.25×1021.44×10−2=8.68×103 N/CE = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

  • From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
  • From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC

6Problem 6medium

❓ Question:

A solid insulating sphere of radius R = 0.08 m has a uniform volume charge density ρ = 6.5 × 10⁻⁶ C/m³. Use Gauss's law to find: (a) the electric field at r = 0.05 m (inside), (b) the electric field at r = 0.12 m (outside), and (c) the total charge of the sphere.

💡 Show Solution

Given:

  • R = 0.08 m
  • ρ = 6.5 × 10⁻⁶ C/m³
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)
  • k = 8.99 × 10⁹ N·m²/C²

Gauss's Law: ∮E⃗·dA⃗ = Q_enc/ε₀

(a) Inside sphere (r = 0.05 m < R):

Choose Gaussian surface: sphere of radius r

By symmetry, E is constant on surface and radial: E(4πr2)=Qencε0E(4πr^2) = \frac{Q_{enc}}{ε_0}

Enclosed charge: Qenc=ρ⋅43πr3Q_{enc} = ρ \cdot \frac{4}{3}πr^3

E=ρr33ε0r2=ρr3ε0E = \frac{ρr^3}{3ε_0r^2} = \frac{ρr}{3ε_0}

E=(6.5×10−6)(0.05)3(8.85×10−12)=3.25×10−72.655×10−11E = \frac{(6.5 \times 10^{-6})(0.05)}{3(8.85 \times 10^{-12})} = \frac{3.25 \times 10^{-7}}{2.655 \times 10^{-11}}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

Or using k = 1/(4πε₀): E=4πkρr3=4π(8.99×109)(6.5×10−6)(0.05)3E = \frac{4πkρr}{3} = \frac{4π(8.99 \times 10^9)(6.5 \times 10^{-6})(0.05)}{3}

E=1.22×104 N/CE = 1.22 \times 10^4 \text{ N/C}

(b) Outside sphere (r = 0.12 m > R):

Total charge enclosed: Q=ρ⋅43πR3=(6.5×10−6)⋅43π(0.08)3Q = ρ \cdot \frac{4}{3}πR^3 = (6.5 \times 10^{-6}) \cdot \frac{4}{3}π(0.08)^3

Q=(6.5×10−6)(2.144×10−3)=1.39×10−8 CQ = (6.5 \times 10^{-6})(2.144 \times 10^{-3}) = 1.39 \times 10^{-8} \text{ C}

Outside, field is like a point charge: E=kQr2=(8.99×109)(1.39×10−8)(0.12)2E = \frac{kQ}{r^2} = \frac{(8.99 \times 10^9)(1.39 \times 10^{-8})}{(0.12)^2}

E=1.25×1021.44×10−2=8.68×103 N/CE = \frac{1.25 \times 10^2}{1.44 \times 10^{-2}} = 8.68 \times 10^3 \text{ N/C}

(c) Total charge:

Already calculated: Q = 1.39 × 10⁻⁸ C = 13.9 nC

Verification: Field at surface using both formulas:

  • From inside: E(R) = ρR/(3ε₀) = 1.96 × 10⁴ N/C
  • From outside: E(R) = kQ/R² = 1.96 × 10⁴ N/C ✓

Answers: (a) E = 1.22 × 10⁴ N/C (b) E = 8.68 × 10³ N/C (c) Q = 13.9 nC

7Problem 7medium

❓ Question:

An infinite cylindrical shell of radius R = 0.05 m carries a uniform surface charge density σ = 3.5 × 10⁻⁶ C/m². Find the electric field: (a) inside the cylinder at r = 0.02 m, (b) outside the cylinder at r = 0.08 m. (c) What is the charge per unit length on the cylinder?

💡 Show Solution

Given:

  • R = 0.05 m
  • σ = 3.5 × 10⁻⁶ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Gauss's Law for cylindrical symmetry:

Choose Gaussian surface: cylinder of radius r and length L

(a) Inside cylinder (r = 0.02 m < R):

No charge enclosed: Qenc=0Q_{enc} = 0

By Gauss's law: E(2πrL)=0E(2πrL) = 0

E=0E = 0

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: Qenc=σ⋅2πRLQ_{enc} = σ \cdot 2πRL

By Gauss's law: E(2πrL)=σ(2πRL)ε0E(2πrL) = \frac{σ(2πRL)}{ε_0}

E=σRε0rE = \frac{σR}{ε_0r}

E=(3.5×10−6)(0.05)(8.85×10−12)(0.08)E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}

E=1.75×10−77.08×10−13=2.47×105 N/CE = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}

(c) Charge per unit length:

λ=QL=σ(2πRL)L=2πRσλ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ

λ=2π(0.05)(3.5×10−6)λ = 2π(0.05)(3.5 \times 10^{-6})

λ=1.10×10−6 C/m=1.10 μC/mλ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}

Alternative formula for outside: E=λ2πε0r=2kλrE = \frac{λ}{2πε_0r} = \frac{2kλ}{r}

E=2(8.99×109)(1.10×10−6)0.08=2.47×105 N/CE = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C} ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

8Problem 8medium

❓ Question:

An infinite cylindrical shell of radius R = 0.05 m carries a uniform surface charge density σ = 3.5 × 10⁻⁶ C/m². Find the electric field: (a) inside the cylinder at r = 0.02 m, (b) outside the cylinder at r = 0.08 m. (c) What is the charge per unit length on the cylinder?

💡 Show Solution

Given:

  • R = 0.05 m
  • σ = 3.5 × 10⁻⁶ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Gauss's Law for cylindrical symmetry:

Choose Gaussian surface: cylinder of radius r and length L

(a) Inside cylinder (r = 0.02 m < R):

No charge enclosed: Qenc=0Q_{enc} = 0

By Gauss's law: E(2πrL)=0E(2πrL) = 0

E=0E = 0

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: Qenc=σ⋅2πRLQ_{enc} = σ \cdot 2πRL

By Gauss's law: E(2πrL)=σ(2πRL)ε0E(2πrL) = \frac{σ(2πRL)}{ε_0}

E=σRε0rE = \frac{σR}{ε_0r}

E=(3.5×10−6)(0.05)(8.85×10−12)(0.08)E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}

E=1.75×10−77.08×10−13=2.47×105 N/CE = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}

(c) Charge per unit length:

λ=QL=σ(2πRL)L=2πRσλ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ

λ=2π(0.05)(3.5×10−6)λ = 2π(0.05)(3.5 \times 10^{-6})

λ=1.10×10−6 C/m=1.10 μC/mλ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}

Alternative formula for outside: E=λ2πε0r=2kλrE = \frac{λ}{2πε_0r} = \frac{2kλ}{r}

E=2(8.99×109)(1.10×10−6)0.08=2.47×105 N/CE = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C} ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

9Problem 9medium

❓ Question:

An infinite cylindrical shell of radius R = 0.05 m carries a uniform surface charge density σ = 3.5 × 10⁻⁶ C/m². Find the electric field: (a) inside the cylinder at r = 0.02 m, (b) outside the cylinder at r = 0.08 m. (c) What is the charge per unit length on the cylinder?

💡 Show Solution

Given:

  • R = 0.05 m
  • σ = 3.5 × 10⁻⁶ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Gauss's Law for cylindrical symmetry:

Choose Gaussian surface: cylinder of radius r and length L

(a) Inside cylinder (r = 0.02 m < R):

No charge enclosed: Qenc=0Q_{enc} = 0

By Gauss's law: E(2πrL)=0E(2πrL) = 0

E=0E = 0

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: Qenc=σ⋅2πRLQ_{enc} = σ \cdot 2πRL

By Gauss's law: E(2πrL)=σ(2πRL)ε0E(2πrL) = \frac{σ(2πRL)}{ε_0}

E=σRε0rE = \frac{σR}{ε_0r}

E=(3.5×10−6)(0.05)(8.85×10−12)(0.08)E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}

E=1.75×10−77.08×10−13=2.47×105 N/CE = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}

(c) Charge per unit length:

λ=QL=σ(2πRL)L=2πRσλ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ

λ=2π(0.05)(3.5×10−6)λ = 2π(0.05)(3.5 \times 10^{-6})

λ=1.10×10−6 C/m=1.10 μC/mλ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}

Alternative formula for outside: E=λ2πε0r=2kλrE = \frac{λ}{2πε_0r} = \frac{2kλ}{r}

E=2(8.99×109)(1.10×10−6)0.08=2.47×105 N/CE = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C} ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

10Problem 10medium

❓ Question:

An infinite cylindrical shell of radius R = 0.05 m carries a uniform surface charge density σ = 3.5 × 10⁻⁶ C/m². Find the electric field: (a) inside the cylinder at r = 0.02 m, (b) outside the cylinder at r = 0.08 m. (c) What is the charge per unit length on the cylinder?

💡 Show Solution

Given:

  • R = 0.05 m
  • σ = 3.5 × 10⁻⁶ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Gauss's Law for cylindrical symmetry:

Choose Gaussian surface: cylinder of radius r and length L

(a) Inside cylinder (r = 0.02 m < R):

No charge enclosed: Qenc=0Q_{enc} = 0

By Gauss's law: E(2πrL)=0E(2πrL) = 0

E=0E = 0

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: Qenc=σ⋅2πRLQ_{enc} = σ \cdot 2πRL

By Gauss's law: E(2πrL)=σ(2πRL)ε0E(2πrL) = \frac{σ(2πRL)}{ε_0}

E=σRε0rE = \frac{σR}{ε_0r}

E=(3.5×10−6)(0.05)(8.85×10−12)(0.08)E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}

E=1.75×10−77.08×10−13=2.47×105 N/CE = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}

(c) Charge per unit length:

λ=QL=σ(2πRL)L=2πRσλ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ

λ=2π(0.05)(3.5×10−6)λ = 2π(0.05)(3.5 \times 10^{-6})

λ=1.10×10−6 C/m=1.10 μC/mλ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}

Alternative formula for outside: E=λ2πε0r=2kλrE = \frac{λ}{2πε_0r} = \frac{2kλ}{r}

E=2(8.99×109)(1.10×10−6)0.08=2.47×105 N/CE = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C} ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

11Problem 11medium

❓ Question:

An infinite cylindrical shell of radius R = 0.05 m carries a uniform surface charge density σ = 3.5 × 10⁻⁶ C/m². Find the electric field: (a) inside the cylinder at r = 0.02 m, (b) outside the cylinder at r = 0.08 m. (c) What is the charge per unit length on the cylinder?

💡 Show Solution

Given:

  • R = 0.05 m
  • σ = 3.5 × 10⁻⁶ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Gauss's Law for cylindrical symmetry:

Choose Gaussian surface: cylinder of radius r and length L

(a) Inside cylinder (r = 0.02 m < R):

No charge enclosed: Qenc=0Q_{enc} = 0

By Gauss's law: E(2πrL)=0E(2πrL) = 0

E=0E = 0

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: Qenc=σ⋅2πRLQ_{enc} = σ \cdot 2πRL

By Gauss's law: E(2πrL)=σ(2πRL)ε0E(2πrL) = \frac{σ(2πRL)}{ε_0}

E=σRε0rE = \frac{σR}{ε_0r}

E=(3.5×10−6)(0.05)(8.85×10−12)(0.08)E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}

E=1.75×10−77.08×10−13=2.47×105 N/CE = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}

(c) Charge per unit length:

λ=QL=σ(2πRL)L=2πRσλ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ

λ=2π(0.05)(3.5×10−6)λ = 2π(0.05)(3.5 \times 10^{-6})

λ=1.10×10−6 C/m=1.10 μC/mλ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}

Alternative formula for outside: E=λ2πε0r=2kλrE = \frac{λ}{2πε_0r} = \frac{2kλ}{r}

E=2(8.99×109)(1.10×10−6)0.08=2.47×105 N/CE = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C} ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

12Problem 12medium

❓ Question:

An infinite cylindrical shell of radius R = 0.05 m carries a uniform surface charge density σ = 3.5 × 10⁻⁶ C/m². Find the electric field: (a) inside the cylinder at r = 0.02 m, (b) outside the cylinder at r = 0.08 m. (c) What is the charge per unit length on the cylinder?

💡 Show Solution

Given:

  • R = 0.05 m
  • σ = 3.5 × 10⁻⁶ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Gauss's Law for cylindrical symmetry:

Choose Gaussian surface: cylinder of radius r and length L

(a) Inside cylinder (r = 0.02 m < R):

No charge enclosed: Qenc=0Q_{enc} = 0

By Gauss's law: E(2πrL)=0E(2πrL) = 0

E=0E = 0

(b) Outside cylinder (r = 0.08 m > R):

Enclosed charge: Qenc=σ⋅2πRLQ_{enc} = σ \cdot 2πRL

By Gauss's law: E(2πrL)=σ(2πRL)ε0E(2πrL) = \frac{σ(2πRL)}{ε_0}

E=σRε0rE = \frac{σR}{ε_0r}

E=(3.5×10−6)(0.05)(8.85×10−12)(0.08)E = \frac{(3.5 \times 10^{-6})(0.05)}{(8.85 \times 10^{-12})(0.08)}

E=1.75×10−77.08×10−13=2.47×105 N/CE = \frac{1.75 \times 10^{-7}}{7.08 \times 10^{-13}} = 2.47 \times 10^5 \text{ N/C}

(c) Charge per unit length:

λ=QL=σ(2πRL)L=2πRσλ = \frac{Q}{L} = \frac{σ(2πRL)}{L} = 2πRσ

λ=2π(0.05)(3.5×10−6)λ = 2π(0.05)(3.5 \times 10^{-6})

λ=1.10×10−6 C/m=1.10 μC/mλ = 1.10 \times 10^{-6} \text{ C/m} = 1.10 \text{ μC/m}

Alternative formula for outside: E=λ2πε0r=2kλrE = \frac{λ}{2πε_0r} = \frac{2kλ}{r}

E=2(8.99×109)(1.10×10−6)0.08=2.47×105 N/CE = \frac{2(8.99 \times 10^9)(1.10 \times 10^{-6})}{0.08} = 2.47 \times 10^5 \text{ N/C} ✓

Answers: (a) E = 0 (inside cylindrical shell) (b) E = 2.47 × 10⁵ N/C (radially outward) (c) λ = 1.10 μC/m

13Problem 13easy

❓ Question:

Two large parallel conducting plates are separated by distance d = 2.0 cm. The plates carry surface charge densities +σ and -σ where σ = 4.5 × 10⁻⁷ C/m². Find: (a) the electric field between the plates, (b) the electric field outside the plates, and (c) the potential difference between the plates.

💡 Show Solution

Given:

  • d = 0.02 m
  • σ = 4.5 × 10⁻⁷ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Field from single infinite sheet: E = σ/(2ε₀)

(a) Field between the plates:

Each plate creates field E = σ/(2ε₀)

  • Positive plate creates field pointing away (to the right)
  • Negative plate creates field pointing toward it (also to the right)

Fields add between plates: Ebetween=σ2ε0+σ2ε0=σε0E_{between} = \frac{σ}{2ε_0} + \frac{σ}{2ε_0} = \frac{σ}{ε_0}

E=4.5×10−78.85×10−12=5.08×104 N/CE = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

  • Positive plate: field to the left
  • Negative plate: field to the right

Fields cancel: Eoutside=σ2ε0−σ2ε0=0E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0

Same reasoning for right side: E = 0

(c) Potential difference:

ΔV=∫0dE dx=EdΔV = \int_0^d E \, dx = Ed

ΔV=(5.08×104)(0.02)=1.016×103 VΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}

ΔV=1.02 kVΔV = 1.02 \text{ kV}

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d E=Vd=Qε0A=σε0E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV

14Problem 14easy

❓ Question:

Two large parallel conducting plates are separated by distance d = 2.0 cm. The plates carry surface charge densities +σ and -σ where σ = 4.5 × 10⁻⁷ C/m². Find: (a) the electric field between the plates, (b) the electric field outside the plates, and (c) the potential difference between the plates.

💡 Show Solution

Given:

  • d = 0.02 m
  • σ = 4.5 × 10⁻⁷ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Field from single infinite sheet: E = σ/(2ε₀)

(a) Field between the plates:

Each plate creates field E = σ/(2ε₀)

  • Positive plate creates field pointing away (to the right)
  • Negative plate creates field pointing toward it (also to the right)

Fields add between plates: Ebetween=σ2ε0+σ2ε0=σε0E_{between} = \frac{σ}{2ε_0} + \frac{σ}{2ε_0} = \frac{σ}{ε_0}

E=4.5×10−78.85×10−12=5.08×104 N/CE = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

  • Positive plate: field to the left
  • Negative plate: field to the right

Fields cancel: Eoutside=σ2ε0−σ2ε0=0E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0

Same reasoning for right side: E = 0

(c) Potential difference:

ΔV=∫0dE dx=EdΔV = \int_0^d E \, dx = Ed

ΔV=(5.08×104)(0.02)=1.016×103 VΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}

ΔV=1.02 kVΔV = 1.02 \text{ kV}

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d E=Vd=Qε0A=σε0E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV

15Problem 15easy

❓ Question:

Two large parallel conducting plates are separated by distance d = 2.0 cm. The plates carry surface charge densities +σ and -σ where σ = 4.5 × 10⁻⁷ C/m². Find: (a) the electric field between the plates, (b) the electric field outside the plates, and (c) the potential difference between the plates.

💡 Show Solution

Given:

  • d = 0.02 m
  • σ = 4.5 × 10⁻⁷ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Field from single infinite sheet: E = σ/(2ε₀)

(a) Field between the plates:

Each plate creates field E = σ/(2ε₀)

  • Positive plate creates field pointing away (to the right)
  • Negative plate creates field pointing toward it (also to the right)

Fields add between plates: Ebetween=σ2ε0+σ2ε0=σε0E_{between} = \frac{σ}{2ε_0} + \frac{σ}{2ε_0} = \frac{σ}{ε_0}

E=4.5×10−78.85×10−12=5.08×104 N/CE = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

  • Positive plate: field to the left
  • Negative plate: field to the right

Fields cancel: Eoutside=σ2ε0−σ2ε0=0E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0

Same reasoning for right side: E = 0

(c) Potential difference:

ΔV=∫0dE dx=EdΔV = \int_0^d E \, dx = Ed

ΔV=(5.08×104)(0.02)=1.016×103 VΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}

ΔV=1.02 kVΔV = 1.02 \text{ kV}

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d E=Vd=Qε0A=σε0E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV

16Problem 16easy

❓ Question:

Two large parallel conducting plates are separated by distance d = 2.0 cm. The plates carry surface charge densities +σ and -σ where σ = 4.5 × 10⁻⁷ C/m². Find: (a) the electric field between the plates, (b) the electric field outside the plates, and (c) the potential difference between the plates.

💡 Show Solution

Given:

  • d = 0.02 m
  • σ = 4.5 × 10⁻⁷ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Field from single infinite sheet: E = σ/(2ε₀)

(a) Field between the plates:

Each plate creates field E = σ/(2ε₀)

  • Positive plate creates field pointing away (to the right)
  • Negative plate creates field pointing toward it (also to the right)

Fields add between plates: Ebetween=σ2ε0+σ2ε0=σε0E_{between} = \frac{σ}{2ε_0} + \frac{σ}{2ε_0} = \frac{σ}{ε_0}

E=4.5×10−78.85×10−12=5.08×104 N/CE = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

  • Positive plate: field to the left
  • Negative plate: field to the right

Fields cancel: Eoutside=σ2ε0−σ2ε0=0E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0

Same reasoning for right side: E = 0

(c) Potential difference:

ΔV=∫0dE dx=EdΔV = \int_0^d E \, dx = Ed

ΔV=(5.08×104)(0.02)=1.016×103 VΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}

ΔV=1.02 kVΔV = 1.02 \text{ kV}

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d E=Vd=Qε0A=σε0E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV

17Problem 17easy

❓ Question:

Two large parallel conducting plates are separated by distance d = 2.0 cm. The plates carry surface charge densities +σ and -σ where σ = 4.5 × 10⁻⁷ C/m². Find: (a) the electric field between the plates, (b) the electric field outside the plates, and (c) the potential difference between the plates.

💡 Show Solution

Given:

  • d = 0.02 m
  • σ = 4.5 × 10⁻⁷ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Field from single infinite sheet: E = σ/(2ε₀)

(a) Field between the plates:

Each plate creates field E = σ/(2ε₀)

  • Positive plate creates field pointing away (to the right)
  • Negative plate creates field pointing toward it (also to the right)

Fields add between plates: Ebetween=σ2ε0+σ2ε0=σε0E_{between} = \frac{σ}{2ε_0} + \frac{σ}{2ε_0} = \frac{σ}{ε_0}

E=4.5×10−78.85×10−12=5.08×104 N/CE = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

  • Positive plate: field to the left
  • Negative plate: field to the right

Fields cancel: Eoutside=σ2ε0−σ2ε0=0E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0

Same reasoning for right side: E = 0

(c) Potential difference:

ΔV=∫0dE dx=EdΔV = \int_0^d E \, dx = Ed

ΔV=(5.08×104)(0.02)=1.016×103 VΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}

ΔV=1.02 kVΔV = 1.02 \text{ kV}

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d E=Vd=Qε0A=σε0E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV

18Problem 18easy

❓ Question:

Two large parallel conducting plates are separated by distance d = 2.0 cm. The plates carry surface charge densities +σ and -σ where σ = 4.5 × 10⁻⁷ C/m². Find: (a) the electric field between the plates, (b) the electric field outside the plates, and (c) the potential difference between the plates.

💡 Show Solution

Given:

  • d = 0.02 m
  • σ = 4.5 × 10⁻⁷ C/m²
  • ε₀ = 8.85 × 10⁻¹² C²/(N·m²)

Field from single infinite sheet: E = σ/(2ε₀)

(a) Field between the plates:

Each plate creates field E = σ/(2ε₀)

  • Positive plate creates field pointing away (to the right)
  • Negative plate creates field pointing toward it (also to the right)

Fields add between plates: Ebetween=σ2ε0+σ2ε0=σε0E_{between} = \frac{σ}{2ε_0} + \frac{σ}{2ε_0} = \frac{σ}{ε_0}

E=4.5×10−78.85×10−12=5.08×104 N/CE = \frac{4.5 \times 10^{-7}}{8.85 \times 10^{-12}} = 5.08 \times 10^4 \text{ N/C}

Direction: from positive to negative plate

(b) Field outside the plates:

On the left of both plates:

  • Positive plate: field to the left
  • Negative plate: field to the right

Fields cancel: Eoutside=σ2ε0−σ2ε0=0E_{outside} = \frac{σ}{2ε_0} - \frac{σ}{2ε_0} = 0

Same reasoning for right side: E = 0

(c) Potential difference:

ΔV=∫0dE dx=EdΔV = \int_0^d E \, dx = Ed

ΔV=(5.08×104)(0.02)=1.016×103 VΔV = (5.08 \times 10^4)(0.02) = 1.016 \times 10^3 \text{ V}

ΔV=1.02 kVΔV = 1.02 \text{ kV}

Note: This configuration is a parallel-plate capacitor!

Capacitance: C = ε₀A/d E=Vd=Qε0A=σε0E = \frac{V}{d} = \frac{Q}{ε_0A} = \frac{σ}{ε_0}

Answers: (a) E = 5.08 × 10⁴ N/C (between plates, uniform) (b) E = 0 (outside plates) (c) ΔV = 1.02 kV

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