The force on a layer of charge

In summary, this difference is valid for E = 0 and E2 = σ / Ɛ0, but Purcell covers a more general case, and I don't see how this difference is fulfilled in other cases.
  • #1
madafo3435
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Homework Statement
I am studying the book Electricity and Magnetism by Edward M. Purcell. I resort to the forum because I have problems in understanding where the equation (1.45) that I indicate at the end of my attached images comes from.
Relevant Equations
.
f0.png
f1.png
I understand that this difference is valid for E = 0 and E2 = σ / Ɛ0, but Purcell covers a more general case, and I don't see how this difference is fulfilled in other cases.

I appreciate the help you can give me.
 
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  • #2
Take a surface of some charge density, and of any general shape. At some point on the surface, construct a very tiny Gaussian pillbox that passes through the surface. At all points on the surface the field due to the surface is orthogonal to the surface, so the flux through the sides of the pillbox approaches zero as we make the pillbox smaller. The net flux out of the pillbox, if the electric fields on either side are ##\vec{E}_1## and ##\vec{E}_2##, will be$$\Phi = \vec{E}_2 \cdot \hat{n} \delta S + \vec{E}_1 \cdot (-\hat{n}) \delta S = (\vec{E}_2 - \vec{E}_1) \cdot \hat{n} \delta S = (E_{n,1} - E_{n,2}) \delta S = \frac{q}{\varepsilon_0} \implies E_{n,1} - E_{n,2} = \frac{\sigma}{\varepsilon_0}$$since the unit outward normals, ##\hat{n}## and ##-\hat{n}##, to the two opposite faces of the pillbox will be oppositely directed.
 
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  • #3
etotheipi said:
Take a surface of some charge density, and of any general shape. At some point on the surface, construct a very tiny Gaussian pillbox that passes through the surface. At all points on the surface the field due to the surface is orthogonal to the surface, so the flux through the sides of the pillbox approaches zero as we make the pillbox smaller. The net flux out of the surface, if the electric fields on either side are ##\vec{E}_1## and ##\vec{E}_2##, will be$$\Phi = \vec{E}_2 \cdot \hat{n} \delta S + \vec{E}_1 \cdot (-\hat{n}) \delta S = (\vec{E}_2 - \vec{E}_1) \cdot \hat{n} \delta S = (E_{n,1} - E_{n,2}) \delta S = \frac{q}{\varepsilon_0} \implies E_{n,1} - E_{n,2} = \frac{\sigma}{\varepsilon_0}$$since the unit outward normals, ##\hat{n}## and ##-\hat{n}##, to the two opposite faces of the pillbox will be oppositely directed.
His explanation was very clear, now I understand. Thank you very much
 
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1. What is the force on a layer of charge?

The force on a layer of charge is the force exerted on a charged surface due to the presence of other charges in its surroundings. This force is a vector quantity and is expressed in units of Newtons (N).

2. How is the force on a layer of charge calculated?

The force on a layer of charge is calculated using Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This can be expressed as F = k*q1*q2/d^2, where F is the force, q1 and q2 are the charges, d is the distance between them, and k is the Coulomb's constant.

3. How does the distance between charges affect the force on a layer of charge?

The distance between charges has an inverse relationship with the force on a layer of charge. This means that as the distance between charges increases, the force decreases, and vice versa. This is because the force is spread out over a larger area as the distance increases, resulting in a weaker force.

4. Can the force on a layer of charge be repulsive?

Yes, the force on a layer of charge can be either attractive or repulsive, depending on the charges involved. Like charges (positive to positive or negative to negative) will repel each other, while opposite charges (positive to negative) will attract each other.

5. How does the size of the charges affect the force on a layer of charge?

The size of the charges has a direct relationship with the force on a layer of charge. This means that as the size of the charges increases, the force also increases, and vice versa. This is because larger charges have a greater amount of electric field, resulting in a stronger force on the layer of charge.

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