Why Does the Electric Field at Point P on a Hemispherical Boss Equal 3E?

[joker_900]

Homework Statement

A large and otherwise plane conducting sheet has on it a hemispherical conducting boss. It is placed in an electric field which, in the absence of theboss, would be uniform, normal to the sheet and of magnitude E. Show that the electric field at point P (the highest point on the boss from the sheet) is 3E.

The Attempt at a Solution

Well I think you're supposed to use a dipole field for this, as I know a sphere in a uniform field develops an electric dipole moment and this seems similar. But I don't really know how to progress. I don't really want a full solution, I would just be grateful if someone gives me a start. Thanks

 

[anathema]

for the interesting problem!



You are correct in thinking that a dipole field can be used to solve this problem. First, we need to consider the electric field at point P due to the hemispherical boss alone. The hemispherical boss can be modeled as a point charge at its center, with a charge of -Q (assuming the sheet has a positive charge). The electric field at point P due to this point charge can be calculated using Coulomb's law: E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance from the point charge to point P.

Next, we need to consider the electric field at point P due to the conducting sheet. The sheet can be modeled as an infinite plane of charge, with a surface charge density of σ (assuming the sheet has a positive charge). The electric field at point P due to this infinite plane of charge can be calculated using the formula: E = σ/2ε0, where ε0 is the permittivity of free space.

Now, we can use the principle of superposition to find the total electric field at point P. Since the electric field due to the boss and the sheet are in opposite directions, we can simply subtract the magnitude of the electric field due to the sheet from the magnitude of the electric field due to the boss to get the total electric field at point P. This can be expressed as: Etotal = |Eboss| - |Esheet| = kQ/r^2 - σ/2ε0.

Substituting the values for Q and σ, we get: Etotal = k(-Q)/r^2 - (σ/2ε0) = -kQ/r^2 - (σ/2ε0). Since we are assuming the sheet has a positive charge, σ will be positive and the total electric field will be negative.

To find the magnitude of the electric field at point P, we can use the Pythagorean theorem to find the distance from the center of the hemispherical boss to point P. This distance can be expressed as: √(r^2 + (r/2)^2) = √(5r^2/4). Substituting this value into the equation for Etotal, we get: Etotal = -kQ/(5r^2/4) - (σ/2

 

[Moetasim]

I would first start by visualizing the situation described in the question. We have a large sheet of conducting material with a hemispherical conducting boss on it, placed in a uniform electric field. The field is normal to the sheet and has a magnitude of E. We are interested in finding the electric field at the highest point (point P) on the boss.

Next, I would think about the properties of electric fields and how they behave around conducting materials. In this case, we know that the electric field at the surface of a conductor is always perpendicular to the surface. Additionally, the electric field inside a conductor is zero. These concepts will be important in solving this problem.

To start, we can consider the electric field at point P as a combination of two fields - the field from the uniform electric field E and the field from the conducting sheet and boss. We can then break down the problem into two parts:

1. Finding the electric field from the uniform field E at point P.
2. Finding the electric field from the conducting sheet and boss at point P.

For the first part, we can use the fact that the electric field from a uniform field is constant and always perpendicular to the surface. Therefore, the electric field at point P due to the uniform field E will also be E and will be directed perpendicular to the surface of the conducting sheet.

For the second part, we can use the concept of image charges. Since the boss is conducting, it will create an image charge on the opposite side of the sheet. This image charge will have the same magnitude but opposite sign as the original charge. By considering the electric field from both the conducting sheet and its image charge, we can find the electric field at point P.

Using these two parts, we can then add the electric fields from the uniform field and the conducting sheet and boss to find the total electric field at point P. This will give us a final value of 3E, as desired.

In conclusion, as a scientist, I would approach this problem by visualizing the situation, considering the properties of electric fields, and breaking the problem down into smaller parts. By using the concepts of uniform fields and image charges, we can find the electric field at point P to be 3E.