Parallel and Spherical Plate Capacitor

The two plates of a parallel plate capacitor are $5 \mathrm{mm}$ apart. A slab of a dielectric, of thickness $4 \mathrm{mm}$ is introduced between the plates with its faces parallel to them. The distance between the plates is adjusted so that the capacitance of the capacitor becomes equal to its original value. If the new distance between the plates equal $8 \mathrm{mm}$, what is the dielectric constant of the dielectric used?

On charging a parallel plate capacitor to a potential V, the spacing between the plates is halved, and a dielectric medium of  ${\epsilon }_r=10$  is introduced between the plates, without disconnecting the d.c. source. Explain, using suitable expressions, how the
(i) capacitance,
(ii) potential difference and
​(iii) energy stored in the capacitor change.

If two similar plates, each of area $A$ having surface charge densities  $+\sigma \text{and}-\sigma$ are separated by a distance $d$ in air, write expressions for the potential difference between the plates.

A parallel plate capacitor with air between the plates has a capacitance C.  Which one is the new capacitance if the distance between the plates is doubled?
 

A parallel plate air capacitor has rectangular plates each of area $20 {\mathrm{cm}}^2$ are separated by a distance of $2 \mathrm{mm}$. The potential difference between the plates is $500$ volt. Calculate capacitance of capacitor.

A parallel plate air capacitor has rectangular plates each of area $20 {\mathrm{cm}}^2$ are separated by a distance of $2 \mathrm{mm}$. The potential difference between the plates is $500$ volt. Calculate charge on each plate.

A parallel plate air capacitor has rectangular plates each of area $20 {\mathrm{cm}}^2$ are separated by a distance of $2 \mathrm{mm}$. The potential difference between the plates is $500$ volt. Calculate electric field intensity between the two plates.

A parallel plate air condenser has a capacity of $20 \mathrm{\mu F}$. What will be the new capacity (in $\mathrm{\mu F}$), if the distance between two plates is doubled?
 

A parallel plate capacitor has plates each of length $20.0 \mathrm{cm}$ and breadth $10.0 \mathrm{cm}$. The separation between the plates is $1 \mathrm{mm}$. Calculate the potential difference between the plates if $1 \mathrm{nC}$ charge is given to capacitor. Calculate electric field between the plates. With the same charge of $1 \mathrm{nC}$, if the separation between plates is doubled, what is the new potential difference?

A parallel plate capacitor is of area $6\mathrm{ }{\mathrm{cm}}^2$ and a separation $3\mathrm{ }\mathrm{mm}$. The gap is filled with three dielectric materials of equal thickness (see figure) with dielectric constant $K_1=10, K_2=12$ and $K_3=14.$ The dielectric constant of a material which when fully inserted in above capacitor, gives same capacitance would be:

Two parallel plate capacitors are constructed one by a pair of iron plates and the second by a pair of copper plates of same area and same spacing. Then -

Bending of electric field lines near the edges of the capacitor plate is called the 'fringing of field'. The cause behind it is:

A voltage is applied between the plates of a parallel plate capacitor filled with a dielectric between the plates. Considering the fringing of field, which of the following statement is wrong?

What do you mean by fringing of a field?

A $600 \mathrm{\mu F}$ capacitor is charged by $250 V$ battery. The capacitor is disconnected from battery and connected to another $600 \mathrm{\mu F}$ capacitor. Energy of the system in joule is 

A $600 \mu F$ capacitor is charged by $250 V$ battery. Magnitude of electrostatic energy stored in capacitor in joule is 

What is the electric field between the plates of a charged parallel plate capacitor with surface charge density ${\mathrm{\sigma }}_0$ on plates?

On assuming earth as a spherical conductor what will be its capacitance?

If area of one plate of a parallel plate capacitor is halved then can the device be called a capacitor?

The radii of two spherical conductors are in ratio $1:2$. The ratio of their capacitance is.