Help Solving a Problem Involving Gaussian & Spherical Balloon Charge

In summary, the electric field of a sphere increases linearly with time, and the radius at any given time is proportional to the time since the sphere's radius was increased linearly.
  • #1
gazepdapi1
54
0
I can't figure out how to get started on this problem. Maybe you guys can help. A spherical balloon carries a total cahrge Q(1C) uniformly distributed over the surface. At t=0, r=r(sub 0), at t=30s, r=2r(sub 0). Find what is the electron field at 90 s at a distance of .5m from the balloon surface. I have to use Gaussean for this problem. I know that you have to get an expression for E(t), the electric field as a function of time. Can you guys help?
thanks
nertil
 
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  • #2
Welcome to the forums,

For future reference, could you please post all homework questions in the appropriate forum, thanks.

With respect to your current question, can you write down a mathematical expression of Guass' law? Can you also write an expression for the area A of the balloon at time t?
 
  • #3
Gausses law
E=kQr/R^3, for a sphere where r is the radius of the imaginary Gauss sphere in the balloon and R is the radius of the ballon. The Surface area of the balloon as a function of t is SA(t) = 4(pi)r^2t. I don't know if that's correct or not. And sorry about the post location. I didn't realize.
thanks
 
  • #4
Ahh, you've skipped a few stages :approve:; now that we have the electric field of a sphere, all we need to consider now is how the radius changes with time.
 
  • #5
thats where I am stuck. I don't know how to relate the two.
 
  • #6
Okay, if we assume that the radius of the balloon increases linearly then we can say that r(t) = r0+ar0t = r0(1+at) , where a is some positive constant. From the conditions we have;

[tex]r(30)=r_{0}\left(1+30a\right) = 2r_{0} \Rightarrow 1+30a = 2\Rightarrow a = \frac{1}{30}[/tex]

[tex]\therefore r(t) = r_{0}\left(1+\frac{t}{30}\right)[/tex]

Can you go from here?
 

FAQ: Help Solving a Problem Involving Gaussian & Spherical Balloon Charge

How do I calculate the electric field of a charged Gaussian balloon?

To calculate the electric field of a charged Gaussian balloon, you will need to use the Gauss's Law formula: E = Q/(4πε₀r²), where E is the electric field, Q is the charge of the balloon, ε₀ is the permittivity of free space, and r is the distance from the center of the balloon. Simply plug in the values and solve for E.

What is the relationship between the electric field and the charge of a spherical balloon?

The electric field of a spherical balloon is directly proportional to its charge. This means that as the charge of the balloon increases, the electric field around it also increases. This relationship is described by the Gauss's Law formula: E = Q/(4πε₀r²).

How does the radius of a Gaussian balloon affect the electric field?

The electric field of a Gaussian balloon decreases as the radius of the balloon increases. This is because the electric field follows an inverse square relationship with the radius, meaning that the electric field decreases exponentially as the distance from the center of the balloon increases. This relationship is described by the Gauss's Law formula: E = Q/(4πε₀r²).

Can I use Gauss's Law to calculate the electric field of a charged spherical balloon?

Yes, you can use Gauss's Law to calculate the electric field of a charged spherical balloon. This law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. The surface in this case would be a sphere with the balloon at its center, and the charge enclosed would be the charge of the balloon.

How can I use the electric field of a Gaussian balloon to solve a problem?

The electric field of a Gaussian balloon can be used to solve a variety of problems, such as calculating the force on a charged particle near the balloon or determining the strength of the electric field at a specific point. Additionally, understanding the relationship between the electric field and the charge of the balloon can help in predicting the behavior of the balloon in different situations.

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