Knoll (knob/bump) on the plate of the capacitor

AI Thread Summary
The discussion focuses on calculating the electric field at the top and base of a hemispherical knoll on a flat charged capacitor plate using the principle of superposition. The electric field away from the knoll is denoted as E_0. At the top of the knoll, the electric field is found to be 3E_0 due to the superposition of the original field and the dipole field created by the knoll. Conversely, at the base of the knoll, the electric field is zero, as the dipole field cancels out the original field. The calculations involve modeling the knoll as a dipole and applying relevant equations to determine the resultant fields.
sergiokapone
Messages
306
Reaction score
17

Homework Statement


The inner surface of one of the plates
flat charged capacitor has a small hemispherical knoll.
Away from it the electric field in the capacitor is equal to ## E_0 ##. Using the principle of superposition find the field at the top and at the base of the knoll.

Homework Equations


Field near metall surface ##E = \frac{\sigma}{\epsilon_0}## (SI units)

The Attempt at a Solution


The idea is to represent plate with knoll as two different things. May be as a plate with hole with the inserted in it sphere, or as a plate with hemisphere lying thereon. Need some help.
 
Physics news on Phys.org
Is this your problem?
 
need picture.
if the knoll protrudes from the plate then the bottom is part of the metalization and the E field there = 0.
 
I found solution. Let's start with the model:
403a07446e9060338a7f7a080574c4c4.png


Dipole momentum of the sphere

\begin{equation}
p = \frac{3}{4\pi}VE_0 = r^3 E_0\label{p}.
\end{equation}

Field of the dipole in general
\begin{equation}
\vec E = \frac{3(\vec p\vec r)}{r^5}\vec r - \frac{\vec p}{r^3}. \label{dipE}
\end{equation}

Field of the dipole at the top
\begin{equation}
\vec E_\text{dip} = \frac{2\vec p}{r^3} = 2\vec E_0. \label{dipEup}
\end{equation}

Due to supperposition principle
\begin{equation}
\vec E = \vec E_0 + \vec E_\text{dip} = 3\vec E_0.
\end{equation}

Field of the dipole at the base
\begin{equation}
\vec E = - \frac{\vec p}{r^3}. \label{dipEbase} = - \vec E_0
\end{equation}

Due to supperposition principle
\begin{equation}
\vec E = \vec E_0 + \vec E_\text{dip} = 0.
\end{equation}
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Back
Top