Doubt on an EM problem regarding gauss law

In summary, the conversation discusses a problem in the book "Introduction to electrodynamics" by Griffith, which involves finding the value of the electric field in a region where two spheres with different charge densities overlap. The person was able to solve the problem, but had a question about why the field in the region of overlap cannot be zero. It is explained that the field must be constant due to symmetries and that the value of the electric field cannot be assumed to be zero even if the integral of the electric field over a surface is zero.
  • #1
ubergewehr273
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There's this problem 2.18 in the book "Introduction to electrodynamics" by Griffith.
The problem says the following,
"Two spheres, each of radius R and carrying uniform charge densities ##+\rho## and ##-\rho##, respectively, are placed so that they partially overlap (Image_01). Call the vector from the positive center to the negative center d. Show that the field in the region of overlap is constant, and find its value."

Well, I was able to solve the problem as expected from the book, however, I wondered why the field in the region of overlap has to be a non-zero quantity. I could very well take a spherical Gaussian surface that resides inside the region of overlap, and since the net charge enclosed in this region is 0, hence the electric field ought to be zero (refer Image_02). Where am I going wrong over here?

PFA the corresponding images.
 

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  • #2
##\int \mathbf E \cdot d \mathbf a## is a surface integral. You can "take outside" the absolute value of E only under particular condition of symmetries (pay attention to the dot product between ## \mathbf E## and the surface element ##d \mathbf a##). For example, in spherical coordinates the field inside the surface has to be radial. Here it is most certainly not radial: it is constant! It is explained in the book that you have to pay attention to symmetries when using gauss law.

##\int \mathbf E \cdot d \mathbf a = 0## does not necessarily imply ##\mathbf E = 0##. Take for example a point charge and apply gauss law to a surface that does NOT contain the charge. According to you reasoning you would be tempted to say that the field inside the surface is zero while it is obviously not zero.
 
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  • #3
dRic2 said:
∫E⋅da=0\int \mathbf E \cdot d \mathbf a = 0 does not necessarily imply E=0\mathbf E = 0. Take for example a point charge and apply gauss law to a surface that does NOT contain the charge. According to you reasoning you would be tempted to say that the field inside the surface is zero while it is obviously not zero.
Thanks for the insight :)
 
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1. What is Gauss's Law and how does it relate to electromagnetism?

Gauss's Law is one of the four Maxwell's equations that describe the behavior of electric and magnetic fields. It states that the electric flux through a closed surface is directly proportional to the charge enclosed by that surface. In other words, it relates the distribution of electric charges to the resulting electric field.

2. How do I apply Gauss's Law to solve an electromagnetic problem?

To apply Gauss's Law, you first need to define a closed surface that encloses the charge or charges of interest. Then, you need to calculate the electric flux through that surface, which can be done by finding the dot product of the electric field and the surface area vector. Finally, you can use the equation Q = ε0ΦE to solve for the charge enclosed, where Q is the total charge, ε0 is the permittivity of free space, and ΦE is the electric flux.

3. Can Gauss's Law be used for any type of charge distribution?

Yes, Gauss's Law can be used for any type of charge distribution, as long as the charge is enclosed by the surface being considered. However, it is most commonly used for symmetric charge distributions, as it simplifies the calculation of the electric flux.

4. What are some common mistakes to avoid when applying Gauss's Law?

One common mistake is not properly defining the closed surface or not considering all charges enclosed by the surface. Another mistake is not taking into account the direction of the electric field, as the electric flux is a vector quantity. Additionally, it is important to use consistent units when plugging values into the equation.

5. How does Gauss's Law relate to the conservation of charge?

Gauss's Law is closely related to the conservation of charge, as it states that the total electric flux through a closed surface is equal to the total charge enclosed by that surface. This means that the amount of charge entering a closed surface must be equal to the amount of charge leaving that surface, thus satisfying the principle of conservation of charge.

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