En anglais

Letting symmetry do the integral

Science · Electricity & Magnetism · ages 18-19

Name ______________________ Date ____________

A spherical object with greater charge per volume in its upper half than in its lower half has spherical charge symmetry.
Circle one: True False
When can Gauss's law give E directly by a surface-area calculation?
- Whenever a closed surface encloses a known total charge
- When symmetry fixes E on faces that carry the flux
- Whenever the Gaussian surface has a familiar geometric shape
What should you check first when planning a symmetry-based Gauss's law solution?
- Choose a convenient surface before checking the charge
- Treat all enclosed charge as a point charge immediately
- Identify the symmetries of the charge distribution
An isolated solid sphere in vacuum has radius R and uniform positive volume charge density. If its surface field is E(R), what is the field magnitude at r=R/3?
- E(R)/9, because field scales with squared radius
- 3E(R), because all of Q acts as a central point
- E(R)/3, because Qenc and area scale differently
An isolated solid sphere in vacuum has radius R and uniform positive volume charge density. If its surface field is E(R), what is its field at r=3R?
- E(R)/9, with the field pointing radially outward
- E(R)/3, with the field pointing radially outward
- 9E(R), with the field pointing radially outward
An isolated infinite line in vacuum has uniform positive charge per length λ. What is its field at perpendicular distance r>0?
- E = λ/(4πε0 r²), pointing away from the line
- E = λ/(2πε0 r), pointing away from the line
- E = λr/(2πε0), pointing away from the line
An isolated infinite thin sheet in vacuum has uniform positive charge per area σ. What field follows from a pillbox with one cap of area A on each side?
- E = σ/ε0, with the field directed away on each side
- E = σ/(4ε0), with the field directed away on each side
- E = σ/(2ε0), with the field directed away on each side
A closed Gaussian surface around an uneven charge distribution encloses zero net charge. What does Gauss's law tell you?
- Net flux is zero; E can still vary over the surface
- E is zero at every point on the Gaussian surface
- Each small surface patch has exactly zero electric flux

Answer key

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Letting symmetry do the integral W1-mt_1LZboKPJ4D-s1

False · Spherical charge symmetry requires charge density to be the same in every direction at a fixed distance from the center.
When symmetry fixes E on faces that carry the flux · Knowing enclosed charge gives net flux. To find E directly, symmetry must also fix the field pattern on the surface.
Identify the symmetries of the charge distribution · Check the charge symmetry before choosing a surface. That symmetry must justify the field pattern used in the flux calculation.
E(R)/3, because Qenc and area scale differently · Inside the uniform sphere, enclosed charge grows as r³ while Gaussian area grows as r². Their ratio makes E proportional to r.
E(R)/9, with the field pointing radially outward · Outside the sphere, all of Q is enclosed and E falls as 1/r². Tripling the radius gives one ninth of the surface field.
E = λ/(2πε0 r), pointing away from the line · A coaxial cylinder has side area 2πrL and encloses charge λL. Its end caps carry no flux, giving E=λ/(2πε0 r).
E = σ/(2ε0), with the field directed away on each side · Each cap contributes EA and the side contributes zero. Enclosed charge is σA, so 2EA=σA/ε0 gives E=σ/(2ε0).
Net flux is zero; E can still vary over the surface · Gauss's law fixes net flux, not each local field value. Positive and negative flux contributions can cancel even when E is not zero.