Two point charges exert force of f newton when separated by a distance of 10 cm in air. The distance for which the force between them will be same if the medium between them has epsilon =4
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In (a) the thickness of the dielectric is the same as the distance between the charges, rr, so the effective thickness is rK−−√rK. If we put this in the force law we get:
F=14πϵ0Q1Q2(rK−−√)2=14πϵ01KQ1Q2r2
F=14πϵ0Q1Q2(rK)2=14πϵ01KQ1Q2r2
as we expect. The force is reduced by a factor of KK.
Now consider (b). To get the effective distance between the charges we have to add the distance through the air, r−tr−t, plus the effective thickness of the dielectric, tK−−√tK, so the effective distance between the charges is:
d=(r−t)+tK−−√
d=(r−t)+tK
and the force is just:
F=14πϵ0Q1Q2d2=14πϵ0Q1Q2(r−t+tK−−√)2
F=14πϵ0Q1Q2d2=14πϵ0Q1Q2(r−t+tK)2
This is how you get the force when the space between the charges is only partially filled by the dielectric.
F=14πϵ0Q1Q2(rK−−√)2=14πϵ01KQ1Q2r2
F=14πϵ0Q1Q2(rK)2=14πϵ01KQ1Q2r2
as we expect. The force is reduced by a factor of KK.
Now consider (b). To get the effective distance between the charges we have to add the distance through the air, r−tr−t, plus the effective thickness of the dielectric, tK−−√tK, so the effective distance between the charges is:
d=(r−t)+tK−−√
d=(r−t)+tK
and the force is just:
F=14πϵ0Q1Q2d2=14πϵ0Q1Q2(r−t+tK−−√)2
F=14πϵ0Q1Q2d2=14πϵ0Q1Q2(r−t+tK)2
This is how you get the force when the space between the charges is only partially filled by the dielectric.
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