Force acting on a dielectric placed between parallel plates

[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?

 

[berkeman]

inserted to a distance x

inserted how? From which direction?

if the capacitor has a charge q

Before or after the dielectric is inserted?

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.

 

[haruspex]

q^2 d(k-1)÷(2εwl^2)

Assuming the insertion is in the length direction, that looks right to me.

 

[carlyn medona]

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)

 

[haruspex]

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.

found force by differentiating energy with respect to distance

That's the method I used, getting the same result.

 

[carlyn medona]

Length direction, and thanks for confirming