Charge On A Spherical Surface

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Homework Statement



Electric charge resides on a spherical surface of radius 0.3 centered at the origin with charge density specified in spherical polar coordinates by [itex]f(r,\phi, \theta) = 3 × 10^{-12} cos(\theta)[/itex].

Determine the total amount of electric charge on the sphere.

Homework Equations



Total charge, [itex]Q = \int\limits_s f dA[/itex]

The Attempt at a Solution



[/B]Essentially, I am confused about why I am finding the answer is zero. What is the physical explanation for this? (provided my maths is ok)

Total charge, [itex]Q = \int\limits_s f dA[/itex]

[itex]Q = \int\limits_0^{2\pi} \int\limits_0^\pi 3 × 10^{-12} cos(\theta) r^{2} \sin(\theta) d\phi d\theta[/itex]

[itex]Q = \int\limits_0^{2\pi} \int\limits_0^\pi 2.7×10^{-13} cos(\theta) \sin(\theta) d\phi d\theta[/itex]

[itex] \int\limits_0^{2\pi} d\phi = 2\pi[/itex]

[itex]Q = 5.4×10^{-13} \pi \int\limits_0^\pi cos(\theta) \sin(\theta) d\theta[/itex]

[itex]\sin(2 \theta) = 2 \sin(\theta) \cos(\theta)[/itex]

[itex]\frac{1}{2} \sin(2 \theta) = \sin(\theta) \cos(\theta)[/itex]

[itex]Q = 5.4×10^{-13} \pi \int\limits_0^\pi \frac{1}{2} \sin(2 \theta) d\theta[/itex]

[itex]Q = 5.4×10^{-13} \pi ((-\frac{1}{4} \cos(2 \pi)) - (-\frac{1}{4}\cos(0)) [/itex]

[itex]Q = 0[/itex]

I can't see an obvious mistake in my maths, but the answer doesn't seem right either.

I really appreciate any help you can give,

thanks!
 

Answers and Replies

  • #2
ehild
Homework Helper
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cos(θ) changes sign at θ=pi/2. The upper hemisphere is positive, the lower one is negatively charged.
 
  • #3
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cos(θ) changes sign at θ=pi/2. The upper hemisphere is positive, the lower one is negatively charged.
Ohhh!

That makes perfect sense, thank you!
 

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