Capacitors,dielectrics and electrostatic forces

AI Thread Summary
When a dielectric is inserted into a capacitor, the electrostatic force acts inward due to the fringing electric field, which is non-uniform. This inward force results from the interaction between the charges on the capacitor plates and the induced charges on the dielectric. Conversely, when the dielectric is removed, the force acts outward, opposing the insertion. The net horizontal force on the dielectric remains zero because equal forces act on both sides. Understanding these forces involves applying energy calculations related to capacitance and distance.
harshasunder
Messages
9
Reaction score
0
hi

in what direction does the electrostatic force act when we insert a dielectric within a capacitor? and when we remove the dielectric? supposedly they act in opposite directions in each case, and either push or pull the dielectric out or into the capacitor. i m not able to understand how the direction is acting perpendicular to the direction in which the charges are. for example, q is the charge on one plate.put a dielectric in and charge on dielectric is -q', where q' is less than q.then force is given by coulombs law, and let it be f. now that f is pulling the dielectric towards the plate no? (attractive force). so where does the force where that pushes or pulls the dielectric out or in come from?
thanks
harsha
 
Last edited:
Physics news on Phys.org
Firstly, the net horizontal force on the dielectric is going to be zero, since there are TWO forces pulling the dielectric to the side, one on each side of the dielectric.

The reason the dielectric is pulled inward is due to the fringing field of the capacitor, the part of the field that is no longer uniform. You can calculate the force by using the formula for energy, .5*Q^2/C and then differentiating with respect to distance, since F = dU/dx.
 
I think actually you would have to integrate over the distance since F= Ux
 
Susskind (in The Theoretical Minimum, volume 1, pages 203-205) writes the Lagrangian for the magnetic field as ##L=\frac m 2(\dot x^2+\dot y^2 + \dot z^2)+ \frac e c (\dot x A_x +\dot y A_y +\dot z A_z)## and then calculates ##\dot p_x =ma_x + \frac e c \frac d {dt} A_x=ma_x + \frac e c(\frac {\partial A_x} {\partial x}\dot x + \frac {\partial A_x} {\partial y}\dot y + \frac {\partial A_x} {\partial z}\dot z)##. I have problems with the last step. I might have written ##\frac {dA_x} {dt}...
Thread 'Griffith, Electrodynamics, 4th Edition, Example 4.8. (Second part)'
I am reading the Griffith, Electrodynamics book, 4th edition, Example 4.8. I want to understand some issues more correctly. It's a little bit difficult to understand now. > Example 4.8. Suppose the entire region below the plane ##z=0## in Fig. 4.28 is filled with uniform linear dielectric material of susceptibility ##\chi_e##. Calculate the force on a point charge ##q## situated a distance ##d## above the origin. In the page 196, in the first paragraph, the author argues as follows ...
Back
Top