Gauss' theorem — Practice Quiz
ICSE · Class 12 · Physics
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Quick Quiz: Gauss' theorem
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A point charge of 4.0 μC is enclosed by a spherical Gaussian surface in vacuum. Find the electric flux through the surface.
A uniformly charged spherical shell has radius 0.20 m and total charge 3.2 μC. Find the electric field just outside its surface and the surface charge density σ.
A point charge creates an electric flux of -2.0 × 10^3 N·m²·C^-1 through a Gaussian sphere. Find the charge enclosed, and state the flux if the sphere radius is doubled while keeping the same enclosed charge.
An infinitely long wire has linear charge density 5.0 × 10^-8 C·m^-1. Find the electric field at a distance 0.20 m from the wire.
Sample Questions
A large plane sheet has surface charge density 1.77 × 10^-11 C·m^-2. Find the electric field near the sheet and the field between two such sheets carrying equal and opposite charge densities of this magnitude.
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E near one sheet = 1.0 N·C^-1, field between opposite sheets = 2.0 N·C^-1
For one infinite plane sheet, E = σ/(2ε₀). With σ = 1.77 × 10^-11 C·m^-2, E = 1.77 × 10^-11 /(2 × 8.85 × 10^-12) = 1.0 N·C^-1. For two equal and opposite sheets, the fields add between the sheets, so E = σ/ε₀ = 2.0 N·C^-1.
A conducting sphere of radius 0.30 m carries charge 2.7 μC. Find the electric field just outside the surface and the electric field inside the sphere.
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E just outside = 2.7 × 10^5 N·C^-1, E inside = 0
Just outside a charged conductor, E = σ/ε₀ = q/(4πR²ε₀). With q = 2.7 × 10^-6 C and R = 0.30 m, E = 9.0 × 10^9 × 2.7 × 10^-6 / 0.09 = 2.7 × 10^5 N·C^-1. Inside a charged conducting sphere, the electric field is zero.
A uniformly charged non-conducting sphere has radius 0.50 m and total charge 6.0 μC. Find the electric field at r = 0.20 m from the centre.
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4.32 × 10^4 N·C^-1
Inside a uniformly charged non-conducting sphere, E = (1/4πε₀)(qr/R³). Substituting q = 6.0 × 10^-6 C, r = 0.20 m, R = 0.50 m gives E = 9.0 × 10^9 × 6.0 × 10^-6 × 0.20 / 0.125 = 4.32 × 10^4 N·C^-1.
A cube encloses a charge of 1.5 μC at any position inside it. Find the total electric flux through the cube, and then the flux if a second charge of -0.5 μC is also enclosed.
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Flux = 1.69 × 10^5 N·m²·C^-1, then 1.13 × 10^5 N·m²·C^-1
Net flux through a closed surface is Φ_E = q/ε₀. For 1.5 μC, Φ_E = 1.5 × 10^-6 / 8.85 × 10^-12 ≈ 1.69 × 10^5 N·m²·C^-1. With an added -0.5 μC, net charge becomes 1.0 μC, so flux becomes 1.0 × 10^-6 / 8.85 × 10^-12 ≈ 1.13 × 10^5 N·m²·C^-1.
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