Dielectric and force between charges

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SUMMARY

The discussion centers on the calculation of the force between two charges, q, located at positions x=-d and x=d, when a dielectric with dielectric constant K is inserted between them. The original force is given by the formula F=q²/(4π(ε₀)4d²). Participants explore whether the force remains unchanged despite the presence of the dielectric, concluding that the electric field E remains the same as it is derived from the displacement field D, which is unaffected by the dielectric. The confusion arises regarding the behavior of the electric field at the dielectric boundary.

PREREQUISITES
  • Understanding of Coulomb's Law and electric force calculations
  • Familiarity with electric displacement field (D) and electric field (E)
  • Knowledge of dielectric materials and their properties
  • Basic principles of electrostatics and boundary conditions
NEXT STEPS
  • Study the effects of dielectrics on electric fields in electrostatics
  • Learn about boundary conditions for electric fields at dielectric interfaces
  • Explore the relationship between electric displacement field (D) and electric field (E) in dielectrics
  • Investigate the implications of dielectric constants on force calculations between charges
USEFUL FOR

Students and professionals in physics, particularly those studying electrostatics, as well as anyone interested in the effects of dielectrics on electric forces and fields.

shomey
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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?
 
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shomey said:
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...
 
F=qE and E=D/(K*epsilonzero)
 
pam said:
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...
 
What happens at the dielectric surface/boundary?
Regards,
Reilly Atkinson
 

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