New Capacitance in Dielectric

Q. The figure shows a capacitor having three layers between its plates. Layer $x$ is vacuum, $y$ is conductor and $z$ is a dielectric. Which of the following change(s) will result in increase in capacitance?

1. Replace $x$ by conductor
2. Replace $y$ by dielectric
3. Replace $y$ by condctor
4. Replace $x$ by dielectric

Q. The capacity of capacitor is $2\mu F$. If a medium is placed between its plates, the capacity becomes $12\mu F$. The dielectric constant of the medium will be
[Pb. PMT 2003]

1. 6
2. 5
3. 4
4. 3

Q. I have an insulator with dielectric constant k. If I place it in a region in vacuum with a field E. Then the net field inside the dielectric will be?

1. $Ek$
2. $\frac{E}{k}$
3. $\frac{E}{k^2}$
4. $E{k}^2$

Q.

I have an insulator with dielectric constant k. If I place it in a region in vacuum with a field E, what is the induced field inside the dielectric?

1. $-\frac{E}{k}$

2. $E(1-k)$

3. $E(1-\frac{1}{k})$

4. $E(1+\frac{1}{k})$

Q. A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in

1. Increase in the potential difference across the plate, reduction in stored energy, but no change in the charge on the plates

2. Reduction of charge on the plates and increase of potential difference

   across the plates

3. Decrease in the potential difference across the plates, reduction in the stored energy, but no change in the charge on the plates

4. None of the above

Q.

The capacity of a parallel plate condenser is . When a glass plate is

placed between the plates of the conductor, its potential becomes 1/8thof

the original value. The value of dielectric constant will be

1. 1.6

2. 5

3. 8

4. 40

Q. When a capacitor of capacitance C is filled with a dielectric (i.e. thickness of the dielectric is equal to the spacing between the plates) of dielectric constant k, what is the new capacitance?

1. $\frac{C}{k}$
2. $C{k}^2$
3. $\frac{C}{k^2}$
4. $Ck$

Q. If I place a dielectric of thickness 2cm and relative permittivity 2 in between the plates of a parallel plate capacitor, with area $3c{m}^2$ and plate spacing 4 cm, what is the new capacitance?

1. $2{\epsilon }_0\times {10}^{-2}F$
2. ${\epsilon }_0\times {10}^{-2}F$
3. None of these.
4. $3{\epsilon }_0\times {10}^{-2}F$

Q.

A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates, which results in:

1. None of the above

2. Reduction of charge on the plates and increase of potential across the plates

3. Decrease in the potential difference across the plates reduction in stored energy, but no change in the charge on the plates
   

4. Increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates

   
   

Q.

When a dielectric slab is introduced between the plates of an isolated charged capacitor, it

1. All of the above

2. Increase the capacitance of the capacitor

3. Decreases the electric filed between the plates

4. Decreases the amount of energy stored in the capacitor