How can I correctly solve for dA in Gauss's Law for a sphere?

In summary, the conversation discusses the difficulty in solving for the dA portion of Guss's Law for a sphere with a point charge on the inside. The book states that the integral of 1dA is equal to 4(pi)r^2, but when attempting to integrate it manually, the result is 2(pi^2)r^2. A suggested integral is also provided to help clarify the concept."
  • #1
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Hey, I'm having some trouble solving for the dA portion of Guss's Law for a sphere as the Gaussian surface and a point charge on the inside.

According to my book, Integral(1dA) = 4(pi)r^2

So when I try to integrate it myself I get 2(pi^2)r^2
The integral I solve is [tex]\int[/tex][tex]\int r\phi r\theta[/tex] where theta = [0,2pi] and phi = [0,pi]

Can anyone set me straight please?
 
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  • #2
On the surface of a sphere, the element of area bounded by [itex]d\theta[/itex] and [itex]d\phi[/itex] has area [itex](r d\theta)(r \sin \theta d\phi) = r^2 \sin \theta d\theta d\phi[/itex].
 

1. How do you solve for dA in Gauss's Law?

To solve for dA in Gauss's Law, you can use the formula dA = E * dS * cos(theta), where E is the electric field, dS is the surface area, and theta is the angle between the electric field and the surface normal. Alternatively, you can also use integration to calculate the net electric flux through a closed surface.

2. What is the significance of solving for dA in Gauss's Law?

Solving for dA in Gauss's Law allows us to calculate the electric flux through a closed surface, which is a crucial concept in understanding the behavior of electric fields. It helps us determine the strength and direction of the electric field at a given point, and is used in many practical applications, such as designing electrical circuits and analyzing the behavior of charged particles.

3. Can you give an example of solving for dA in Gauss's Law?

Sure, let's say we have a point charge q located at the center of a spherical surface with radius r. We want to find the electric flux through the surface. By symmetry, we know that the electric field is constant and perpendicular to the surface at every point. Therefore, we can use the simplified formula dA = E * dS, where dS is the surface area of the sphere. Substituting this into Gauss's Law, we get q = E * 4*pi*r^2. Solving for E, we get E = q / (4*pi*epsilon*r^2). This is the electric field at any point on the surface, and we can use it to calculate the electric flux through the entire surface.

4. What is the role of dA in Gauss's Law?

dA, or the differential surface area, is a crucial element in Gauss's Law as it represents the small area element through which the electric flux is calculated. By summing up the contributions of all these small areas, we can determine the total electric flux through a closed surface. Therefore, dA plays a vital role in accurately calculating the electric field and understanding the behavior of electric fields.

5. Are there any limitations to using Gauss's Law to solve for dA?

Yes, there are a few limitations to using Gauss's Law to solve for dA. First, it only applies to static electric fields, so it cannot be used for time-varying fields. Additionally, it assumes that the electric field is uniform and perpendicular to the surface at every point, which may not always be the case. Finally, Gauss's Law is only valid for closed surfaces, so it cannot be used to calculate the electric flux through open surfaces.

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