# Capacitor,dielectric and force on plates

hello i had a question bothering me very much,

A dielectric slab is inserted between the plates of an isolated capacitor.The force between the plates will
a)increase b)decrease
c)remains unchanged d)becomes zero
......
ans given:(c)

I have 2 cases

1)charged with battery, removed battery ,then dielectric placed

so if we place a dielectric the field between the plates fall by a factor K(dielectric constant),so the force should change

2)charged with battery,battery remains,then dielectric placed

so if i consider this case,since the potential difference and distance remain constant from relation V=Ed , field E would remain the same,so as to keep it constant charge on the capacitor is increased by getting additional charge from the battery.

so the force on the plates should increase because though the E does not change as free charge on plates increases...

so force changes in both cases according to me...where i was wrong!!

if the force remain unchanged then how does the energy stored increases than the case where no dielectric is there!!!

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berkeman
Mentor
hello i had a question bothering me very much,

A dielectric slab is inserted between the plates of an isolated capacitor.The force between the plates will
a)increase b)decrease
c)remains unchanged d)becomes zero
......
ans given:(c)

I have 2 cases

1)charged with battery, removed battery ,then dielectric placed

so if we place a dielectric the field between the plates fall by a factor K(dielectric constant),so the force should change

2)charged with battery,battery remains,then dielectric placed

so if i consider this case,since the potential difference and distance remain constant from relation V=Ed , field E would remain the same,so as to keep it constant charge on the capacitor is increased by getting additional charge from the battery.

so the force on the plates should increase because though the E does not change as free charge on plates increases...

so force changes in both cases according to me...where i was wrong!!

if the force remain unchanged then how does the energy stored increases than the case where no dielectric is there!!!

What are the equations for the following:

** charge Q in terms of the capacitance and voltage?

** capacitance C in terms of the geometry of the capacitor and the dielectric constant?

** force F between two charged bodies in terms of their separation (and something related to dielectric constant)?

And what is the difference in the two test cases posed? What changes in one case, but is held constant in the other?

Q=CV
C=(epsilon)*A*K/d (K here dielectric constant)
F=K1*Q1*Q2/d^2 (K1 is constant)
in one case the battery is removed and then the dielectric is placed,
and in the other dielectric placed with still battery connected

berkeman
Mentor
Q=CV
C=(epsilon)*A*K/d (K here dielectric constant)
F=K1*Q1*Q2/d^2 (K1 is constant)
in one case the battery is removed and then the dielectric is placed,
and in the other dielectric placed with still battery connected
In your C= equation, is K $$\epsilon_r$$ ?

In your F= equation, read a bit more about K1 -- what goes into it? Maybe not so constant in this problem.

And what is different about leaving the battery connected? Why would that change something?

yes K is relative permittivity ....
yes K1 isn't a constant it changes in this situation i just didnt notice that thank u....
of course they are two different cases i think because in case of battery disconnected the electric field falls by value of relative permittivity,some charge on the plate leaves to balance this
where as when battery is connected the field should be constant as potential and distance are constant,but we just put in a dielectric so E field has to fall but its constant..so in order to keep it constant additional charge comes on to the plates

berkeman
Mentor
yes K is relative permittivity ....
yes K1 isn't a constant it changes in this situation i just didnt notice that thank u....
of course they are two different cases i think because in case of battery disconnected the electric field falls by value of relative permittivity,some charge on the plate leaves to balance this
where as when battery is connected the field should be constant as potential and distance are constant,but we just put in a dielectric so E field has to fall but its constant..so in order to keep it constant additional charge comes on to the plates
Good, you're getting closer. When the battery is disconnected, the constant factor is the charge on the two plates. When the battery stays connected, the constant is the voltage on the two plates.

k yes the charge should remain the same,its not the charge flowing out..but the force changing because now we have a relative permittivity in the force equation constant!!