Electrostatics problem (Spherical surface)

[ravager1987]

Homework Statement

Find the electric field a distance z above the center of a spherical surface of radius R, which carries a uniform charge density "sigma" Treat the case z<R(inside) as well z>R (outside). express your answers in terms of the total charge q on the sphere.[hint:use the law of cosines to write r interms of R and "theta". be sure to take the positive square root:
sqrt(R^2+z^2-2Rz)=(R-z) if R>z and (z-R) for R<z]


here is the solution for the problem:
http://img689.imageshack.us/img689/2492/questionr.jpg

even with the solution i have problem understanding...
what i don't understand is:
1) how did he get dq=sigma*R^2*sin(theta)d(theta)d(phi)
2)the reason for taking cos(curlyphi) ?
my guess: to get the vertical component? since horizontal components cancels out
3)is there a different way looking at this problem? i remember in college there is a simpler way doing this kind of problem...

 

[Delphi51]

1. R*sinθ*dφ and R*dθ are the length and width of the area da where the charge dq lies.
2. You are correct.
3. Yes, you could use Gauss' Law. This problem must be at a point in your course where you have not taken Gauss' Law - and it is good practise!

 

[kerimek]

Dear student,

Thank you for reaching out for clarification on this electrostatics problem. I will do my best to address your questions and provide additional insights.

1) To understand how dq=sigma*R^2*sin(theta)d(theta)d(phi) was obtained, we need to think about the definition of charge density (sigma). Charge density is defined as the amount of charge per unit volume. In this case, we have a spherical surface, so we can think of it as a collection of infinitesimally small rings stacked on top of each other. The charge on each ring is dq and it is equal to the product of the charge density (sigma) and the volume of the ring. The volume of the ring can be expressed as the circumference of the ring (2*pi*R*sin(theta)) multiplied by the thickness of the ring (R*d(theta)). Therefore, we get dq=sigma*2*pi*R^2*sin(theta)*d(theta). To account for the entire sphere, we need to integrate this expression over all values of theta and phi, which is why we have the additional d(phi) term. This integration gives us the final expression of dq=sigma*R^2*sin(theta)d(theta)d(phi).

2) The reason for taking cos(curlyphi) is to account for the direction of the electric field. As you correctly guessed, we want to find the vertical component of the electric field. By taking cos(curlyphi), we are essentially projecting the electric field vector onto the vertical axis. This is important because the electric field at a point above the center of the sphere will have both horizontal and vertical components. However, we are only interested in the vertical component in this problem.

3) There may be different ways to approach this problem depending on the context and the level of understanding of the student. However, the solution provided is a standard and systematic way of solving this type of problem using the principles of electrostatics. If you remember a simpler way of solving this problem, it would be great to share it with me so I can provide you with a more specific response.

I hope this helps clarify some of your questions. If you need further assistance, please do not hesitate to ask. Keep up the good work in your studies of electrostatics!

Best regards,