How to Calculate Charge Content of a Sphere with Spherical Symmetry?

AI Thread Summary
The discussion focuses on calculating the total charge content of a sphere with a given charge density that varies with radius. The charge density is defined as ρ = ρ₀r/R for 0 ≤ r ≤ R. The initial approach involves using spherical coordinates to express the volume element dV and integrating to find the total charge Q. However, the calculation is identified as incorrect, particularly in the definition of dV, which needs to be corrected for accurate integration. The correct method involves considering spherical shells and integrating the charge over these shells from 0 to R.
Hannisch
Messages
114
Reaction score
0

Homework Statement


A charge distribution with spherical symmetry has the density \rho = \rho _0 r/R for 0≤ r ≤ R. Determine the total charge content of the sphere.

Homework Equations



\rho = Q / V

The Attempt at a Solution



I started by thinking of the charge dQ of a small volume dV, since

\rho = dQ / dV

I used spherical coordinates to define dV, and said that dV would be

dV = (r d \varphi )(r d \theta) dr

Where \varphi goes from 0 to 2*pi, \theta goes from -pi/2 to pi/2, and r goes from 0 to R, thus covering the entire sphere.

So:

dQ = \rho dV = \rho r^2 d \varphi d \theta dr = \frac{\rho _0 r}{R} r^2 d \varphi d \theta dr = \frac{\rho _0 r^3}{R} d \varphi d \theta dr

I then integrated over this as:

Q = \int ^ {2 \pi} _ {0} \int ^ {\pi /2} _ {-\pi /2} \int ^ {R} _ {0} \frac{\rho _0 r^3}{R} d \varphi d \theta dr

Q = 2 \pi (\pi /2 + \pi /2) \frac{\rho _0 }{R} \int ^ {R} _ {0} r^3 dr = 2 \pi ^2 \frac{\rho _0 }{R} \frac{R^4}{4} = \frac{1}{2} \pi ^2 \rho _0 R^3

And this is not correct, and I can't figure out where I've gone wrong. (It's supposed to be only \pi \rho _0 R^3
 
Physics news on Phys.org
consider any spherical shell inside the sphere of radius x(<R) and thicknedd dx
find charge on it and then integrate it dx from 0 to R
 
Hannisch said:
I used spherical coordinates to define dV, and said that dV would be

dV = (r d \varphi )(r d \theta) dr

Where \varphi goes from 0 to 2*pi, \theta goes from -pi/2 to pi/2, and r goes from 0 to R, thus covering the entire sphere.

Check dV. It is wrong.

ehild
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top