# Charge inside a sphere

1. Oct 18, 2015

### goonking

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

so plugging in the variables, the density should be 0.002

with density known, and the volume of the sphere with radius 10, we can find the mass. (density x volume = mass)

it states the amount of charge is symmetric so I assume the number of protons equals to number of electrons.

dividing the mass 16.75 kg by mass of a proton + electron (1.623 x10^-27 kg + 9.11 x 10^-31) = 1.03 x 10^28 pairs of proton and electron

each pair has charge 2q then, so 1.03 x 10^28 x 2(1.602 x10^-19 C) = 3.3 x 10^9 C in the sphere with radius 10.

Is this correct?

Last edited: Oct 18, 2015
2. Oct 18, 2015

### Staff: Mentor

By spherically symmetric they mean that it looks the same from any direction. It does not refer to the types of charges involved or their mixture. Given the density function is always positive for any radius, you can assume that the charge is positive everywhere.

There is no mass involved here. The density is in Coulombs per cubic meter. It's a charge density. This density varies with the radial distance from the center, and has the same profile in all directions since it depends only on radius, not direction (hence "spherically symmetric").

You need to work out how much charge (in Coulombs) is contained within a sphere of radius R/2. It's analogous to finding a mass when you integrate the mass density over a volume, but in this case it's charge density you're working with. Note that the density varies with radius: that's the tricky bit.

3. Oct 18, 2015

### goonking

why do we assume this? Can't it have a overall negative charge?

4. Oct 18, 2015

### Staff: Mentor

The charge constant in the density formula is $C = 0.004 C/m^3$. It's a positive constant. The problem doesn't introduce any other details and doing so arbitrarily would only add unnecessary complexity.

5. Oct 18, 2015

### goonking

so taking out the constants of my integral, i have

(0.002C/m^3) (4π) ∫ r2 dr

my lower and upper bounds are 0 and 10, since the radius of the sphere we are looking at is R/2 = 20/2 = 10

(0.002C/m^3) (4π) r3

(0.002C/m^3) (4π) ((10)3 - (0)3)
I got an answer close to 25.

6. Oct 18, 2015

### Staff: Mentor

I don't see the density function in there. Density is not constant throughout the sphere. It varies with the radius according to the given density function.

7. Oct 18, 2015

### goonking

ok, so it is

C4π ∫(1- (r/20)) (r2) dr

= C4π r( r2 - (r3 /20))

= C4π (r3 - r4/20))

= (.004C/m3) 4π (1000 - 500)

correct?

8. Oct 18, 2015

### Staff: Mentor

Check your integration. I think you're forgetting to divide by the exponent value: $\int x^n dx = \frac{1}{n+1}x^{n+1}$.

9. Oct 18, 2015

### goonking

10. Oct 18, 2015

### goonking

11. Oct 18, 2015

### Staff: Mentor

That still looks off to me. Check again.

12. Oct 18, 2015

### goonking

r(r2/2 - (r3/3)/20)

= (r3/2 - (r4/3)/20)

unless I did the integral wrong

13. Oct 18, 2015

### DeldotB

$\int_{0}^{\frac{R}{2}}4\pi r^2\rho (r)dr\neq 16.755$

14. Oct 18, 2015

### goonking

or is it

∫(1 -r/20)(r2) dr

= ∫ (r2 - r3/20) dr

= ((r2 - r3/20)^2 ) /2

??

15. Oct 18, 2015

### DeldotB

$\int_{0}^{\frac{R}{2}}4\pi r^2\rho (r)dr=4\pi C\int_{0}^{R/2}r^2\left ( 1-\frac{r}{R} \right )dr=4\pi C\int_{0}^{R/2} r^2-\frac{r^3}{R}dr$
I trust you should be able to do the rest

16. Oct 18, 2015

### Staff: Mentor

Yeah, I think you need to take another look at what you're integrating:
$$Q = \int_0^{10} 4\pi r^2 C \left(1 - \frac{r}{20} \right) dr$$
Distribute the $r^2$ first, then integrate.

17. Oct 18, 2015

### goonking

=4πC (103 - 104/80)
= 10.47

18. Oct 18, 2015

### Staff: Mentor

Huzzah! Now add the units to your result, perhaps check that the number of significant figures is good, and it'll be done.

19. Oct 19, 2015

### goonking

Now, considering that same cloud of mysterious charged particles as in the previous problem with the same charge density function, determine the electric field at a point located 48 % of the way to R, (which is still 20 meters). C is also still the same: 0.004 C/m^3

I'm guessing I still need to use an integral for this?

with the bounds 0 to 9.6m (48% of 20 is 9.6)

20. Oct 19, 2015

### DeldotB

Use Gauss's Law