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Dielectric and force between charges

  • Thread starter shomey
  • Start date
  • #1
30
0
Homework Statement

suppose I have a charge q at (x=-d) and a charge q at (x=+d).
the force between them is q^2/(4*pi*(eps_0)*4*d^2).

now, I insert a dielectric (K) between (-d/2<x<d/2), what would be the force between the charges now?



The attempt at a solution

it seems like it would be the same but it sounds strange...
If I use the D field, it is not effected by the dielectrics, and thus I can see that the electrical field E is the same as before (D/eps_0) and thus the force is the same...

could it be?????
 

Answers and Replies

  • #2
30
0
Homework Statement

suppose I have a charge q at (x=-d) and a charge q at (x=+d).
the force between them is q^2/(4*pi*(eps_0)*4*d^2).

now, I insert a dielectric (K) between (-d/2<x<d/2), what would be the force between the charges now?



The attempt at a solution

it seems like it would be the same but it sounds strange...
If I use the D field, it is not effected by the dielectrics, and thus I can see that the electrical field E is the same as before (D/eps_0) and thus the force is the same...

could it be?????

someone?? please???
I really need help with this...
 
  • #3
pam
455
1
F=qE and E=D/(K*epsilonzero)
 
  • #4
30
0
F=qE and E=D/(K*epsilonzero)
that's just it, i don't think this is it.
notice that the two charge are in the matter eps_0.

* the other dielectric is only found at (-d/2 < x < d/2 ).
* the two charges are in (x = -d) and (x = +d).

so it seems like the force will be F=qE and E=d/eps_0.
which is exactly like if there was no dielectric between them, and it seems pretty weird to me...
 
  • #5
reilly
Science Advisor
1,075
1
What happens at the dielectric surface/boundary?
Regards,
Reilly Atkinson
 

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