Force acting on a dielectric placed between parallel plates

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Homework Help Overview

The discussion centers around a problem involving a parallel plate capacitor with a dielectric material inserted between the plates. The original poster seeks to determine the force acting on the dielectric when it is near the edge of the capacitor, given specific parameters such as charge, dielectric constant, and dimensions of the capacitor.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the effective capacitance and force calculations related to the dielectric's insertion. Questions are raised about the direction of insertion and the conditions of the capacitor before and after the dielectric is placed. Some participants request detailed calculations and diagrams to clarify the setup.

Discussion Status

The discussion is active, with participants confirming each other's calculations and clarifying the direction of the dielectric's insertion. There is a shared understanding of the approach to finding the force, although no consensus on the complete solution has been reached.

Contextual Notes

Participants note constraints such as the absence of an external power source and the assumption that the charge remains constant during the dielectric's insertion. There is also mention of difficulties in providing a diagram to illustrate the setup.

carlyn medona

Homework Statement


A parallel plate capacitor of length l and width w and separation d, has a dielectric of dielectric constant k inserted to a distance x, if the capacitor has a charge q, find force acting on the dielectric when its near the edge

Homework Equations

The Attempt at a Solution

. [/B]
I found effective capacitance which was εw(kx+l-x)÷d and found force to be equal to q^2 d(k-1)÷(2εwl^2) is this right?
 
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carlyn medona said:
inserted to a distance x
inserted how? From which direction?
carlyn medona said:
if the capacitor has a charge q
Before or after the dielectric is inserted?
carlyn medona said:
I found effective capacitance which was εw(kx+l-x)÷d and found force to be equal to q^2 d(k-1)÷(2εwl^2)
Can you please show your detailed calculation? And a diagram would be helpful as well. Thanks.
 
carlyn medona said:
q^2 d(k-1)÷(2εwl^2)
Assuming the insertion is in the length direction, that looks right to me.
 
It was inserted from top, can't get the diagram, and there is no external power source so I expect q to be a constant, so I found capacitance εw(kx-x+l)÷d and energy stored q^2d÷(2εw(kx+l-x) so I found force by differentiating energy with respect to distance and got. q^2d(k-1)÷(2εwl^2)
 
carlyn medona said:
It was inserted from top
But is that the "length" direction or the "width" direction? Judging from your reference to l-x it is the length direction.
carlyn medona said:
found force by differentiating energy with respect to distance
That's the method I used, getting the same result.
 
Length direction, and thanks for confirming
 

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