Calculating Charge on a Spherical Surface with Varying Density

[BOAS]

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 $f(r,\phi, \theta) = 3 × 10^{-12} cos(\theta)$.

Determine the total amount of electric charge on the sphere.

Homework Equations

Total charge, $Q = \int\limits_s f dA$

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, $Q = \int\limits_s f dA$

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

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

$\int\limits_0^{2\pi} d\phi = 2\pi$

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

$\sin(2 \theta) = 2 \sin(\theta) \cos(\theta)$

$\frac{1}{2} \sin(2 \theta) = \sin(\theta) \cos(\theta)$

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

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

$Q = 0$

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!

 

[ehild]

cos(θ) changes sign at θ=pi/2. The upper hemisphere is positive, the lower one is negatively charged.

 

[BOAS]

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!