Capacitor,dielectric and force on plates

In summary, the force between two charged bodies in an isolated capacitor remains unchanged when a dielectric slab is inserted between the plates, regardless of whether the battery is connected or disconnected. This is because in both cases, the constant factor in the force equation is the charge on the two plates, not the battery. The difference in the two cases is that when the battery is disconnected, the electric field falls by a factor of the relative permittivity, causing some charge to leave the plates to balance it. When the battery remains connected, the electric field must remain constant, so additional charge is added to the plates to maintain this.
  • #1
bharath423
30
0
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!

please kindly reply,thank you.
 
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  • #2
bharath423 said:
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!

please kindly reply,thank you.

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?
 
  • #3
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
 
  • #4
bharath423 said:
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 [tex]\epsilon_r[/tex] ?

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?
 
  • #5
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
 
  • #6
bharath423 said:
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.
 
  • #7
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!
 

1. What is a capacitor and how does it work?

A capacitor is an electrical component that stores energy in an electric field. It consists of two conductive plates separated by a dielectric material. When a voltage is applied across the plates, the electric field between them is created, allowing the capacitor to store energy. The amount of energy that can be stored is determined by the capacitance, which is the ratio of the charge on each plate to the applied voltage.

2. What is a dielectric and why is it important in a capacitor?

A dielectric is an insulating material that is placed between the two conductive plates of a capacitor. It is important because it increases the capacitance of the capacitor by allowing it to store more charge at a given voltage. This is because the dielectric material reduces the electric field strength between the plates, allowing for more charge to be stored on the plates.

3. How is the capacitance of a capacitor calculated?

The capacitance of a capacitor can be calculated using the formula C = εA/d, where C is the capacitance, ε is the permittivity of the dielectric material, A is the surface area of the plates, and d is the distance between the plates. This formula shows that the capacitance is directly proportional to the area of the plates and inversely proportional to the distance between them.

4. What is the force on the plates of a capacitor?

The force on the plates of a capacitor is the result of the electric field between the plates. The force is directly proportional to the electric field strength and the charge on the plates. The direction of the force is determined by the polarity of the charge on the plates.

5. How does the type of dielectric affect the capacitance of a capacitor?

The type of dielectric used in a capacitor can greatly affect its capacitance. This is because different materials have different permittivity values, which affects the electric field strength and therefore the capacitance. For example, a capacitor with a vacuum as the dielectric will have a higher capacitance than one with air as the dielectric, since the permittivity of a vacuum is lower than air.

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