Just need a check Finding total charge when given distribution

I was actually wondering how to evaluate that integral without using IBP. But yeah, that definitely makes sense.In summary, we have a sphere with radius R and a charge distribution of ρ(r)=ρ0 cos(θ/3) sin(φ/2). The task is to find the total charge in terms of R and p0. After careful integration and a correction of a mistake, the final answer is (Rho)(R^3)(9/4) C. An alternative method that could have been used to find the answer is by applying the identity \sin\theta\cos\left(\phi\right)=\frac{ \sin\left(\theta+\phi)+\sin\left(\theta-\phi\
  • #1
xxbigelxx
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Homework Statement



We have a sphere (radius R) where charge is placed inside of it, ρ(r)=ρ0 cos(θ/3) sin(φ/2). Find the total charge. Your answer will be in terms of R and p0.


Homework Equations





The Attempt at a Solution




I think I did it correctly, but I would just like to make sure I did (and thus ensure my understanding). My answer is marked by the star, and everything below it is just my integration work that was required by the problem.
 

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  • #2
xxbigelxx said:

Homework Statement



We have a sphere (radius R) where charge is placed inside of it, ρ(r)=ρ0 cos(θ/3) sin(φ/2). Find the total charge. Your answer will be in terms of R and p0.


Homework Equations





The Attempt at a Solution




I think I did it correctly, but I would just like to make sure I did (and thus ensure my understanding). My answer is marked by the star, and everything below it is just my integration work that was required by the problem.

I think you'll want to double check your integration of

[tex]\int_0^{\pi}\cos\left(\frac{\theta}{3}\right)\sin\theta d\theta[/tex]
 
  • #3
Hmm I tried to go over it again and I got the same result. Was it somewhere on the first integration by parts where I messed up? Thanks
 
  • #4
Ohhh I was double checking the wrong part. I ended up making a mistake at the very bottom. It changed my result from that integration to 27/16 and my final answer is now...

(Rho)(R^3)(9/4) C

I hope this should be correct now...
 
  • #5
xxbigelxx said:
Ohhh I was double checking the wrong part. I ended up making a mistake at the very bottom. It changed my result from that integration to 27/16 and my final answer is now...

(Rho)(R^3)(9/4) C

I hope this should be correct now...

Looks good to me.

For future reference, you could have avoided IBP by using a trig identity:

[tex]\sin\theta\cos\left(\frac{\theta}{3}\right)=\frac{\sin\left(\frac{4\theta}{3}\right)+\sin\left(\frac{2\theta}{3}\right)}{2}[/tex]
 
  • #6
Ohh ok thanks. I don't even think I have ever come across that identity before.
 
  • #7
xxbigelxx said:
Ohh ok thanks. I don't even think I have ever come across that identity before.

Its just an application of the identity

[tex]\sin\theta\cos\left(\phi\right)=\frac{ \sin\left(\theta+\phi)+\sin\left(\theta-\phi\right)}{2}[/tex]
 
  • #8
Oh ok thanks a lot.
 

1. What is the formula for finding the total charge when given a distribution?

The formula for finding the total charge when given a distribution is Q = ∑ qi, where Q is the total charge and qi is the charge of each individual distribution.

2. How do I determine the charge of each individual distribution?

The charge of each individual distribution can be determined by multiplying the number of units of charge (such as Coulombs) by the number of distributions in that unit. For example, if there are 5 distributions with 2 Coulombs each, the charge of each individual distribution would be 10 Coulombs.

3. Can the distribution of charges be in different units?

Yes, the distribution of charges can be in different units as long as they are converted to the same unit before calculating the total charge. For example, if there are 3 distributions with 5 microCoulombs each and 2 distributions with 10 nanoCoulombs each, the total charge would be calculated by converting the nanoCoulombs to microCoulombs and then adding all the charges together.

4. When do I use positive and negative signs in the distribution of charges?

Positive and negative signs are used to indicate the direction of the charge. Positive charges signify that the charge is moving in the direction of the electric field, while negative charges signify that the charge is moving opposite to the direction of the electric field.

5. Can the distribution of charges be three-dimensional?

Yes, the distribution of charges can be three-dimensional. In this case, the charge of each individual distribution would be represented by a vector with magnitude and direction. The total charge would then be calculated by adding all the individual charges together using vector addition.

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