2011 AP Physics C E&M frq #1 clarification

[cubejunkies]

Problem 1 on last year's normal (not form b) frq for e&m part a asked to use Gauss' law to prove essentially that at any point P within a thin spherical shell of charge density σ = q/A and radius r that the electric field at this ambiguous point P is zero. The solutions for this problem went typically from the surface integral of the dot product of E and dA, and yielding that E(4πr²) = Q_enclosed/ε₀, and then merely states with words that any explanation relating spherical symmetry to the problem gets a point. I tried doing some long messy integrals but only ended up with E = 2σ/ε₀ which is definitely not the answer because although it is independent of the distance from the center of the shell, it's not zero.

Sorry if my question doesn't make sense.

I'm just looking for a more rigorous, thorough explanation for this problem

Thanks!
Anthony

http://apcentral.collegeboard.com/apc/public/repository/ap11_frq_physics_cem.pdf
http://apcentral.collegeboard.com/a..._electricity_magnetism_scoring_guidelines.pdf

 

[jbsteele]

The solution to this problem can be found in the second link above. Essentially, you need to make use of the fact that the electric field is symmetrical about the center of the shell. This means that the electric field at any point on the surface of the shell is equal and opposite to the electric field at any other point on the surface of the shell. Therefore, the electric field at the center of the shell (point P) must be zero. To prove this mathematically, you can start by writing the Gauss' Law equation: ∮E⋅dA = Q/ε0Then, since the charge density is given as σ = q/A, we can substitute this into the equation above: ∮E⋅dA = Q/ε0 = (σA)/ε0Next, we can plug in the area of the spherical shell (4πr2): ∮E⋅dA = (σA)/ε0 = (σ4πr2)/ε0Now, since the electric field is equal and opposite at every point on the surface of the shell, we can simplify this equation to: ∮E⋅dA = (σ4πr2)/ε0 = 4πr2E/ε0 Finally, we can solve this equation for E to get: E = 0 Therefore, the electric field at any point within a thin spherical shell of charge density σ = q/A and radius r is zero.