Parallel 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?
 

How will the voltage $\left(V\right)$ between the two plates of a parallel plate capacitor depend on the distance $\left(d\right)$ between the plates, if the charge on the capacitor remains the same?

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 -

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 

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

The radius of plate used in a parallel plate capacitor is $10 \mathrm{cm}$. If the plate separation is $10 \mathrm{cm}$ determine the capacitance of this capacitor for air medium.

A capacitor is half-filled with a dielectric material $\left(K=2\right)$ as shown in diagram $\left(1\right)$. If the same dielectric material is to be filled in the capacitor as shown in diagram $\left(2\right)$, then what would be the thickness of the dielectric so that the capacitor still has the same capacitance?

A capacitor is connected by a battery of emf $\mathrm{ }{\mathrm{V}}_0\mathrm{ }$ . The plates of capacitor are lowered into a vessel containing a dielectric liquid with constant velocity $\mathrm{v}$ . Then -

A parallel plate capacitor is connected by a battery of emf $\mathrm{ }{\mathrm{V}}_0$. The plates of capacitor are lowered into a vessel containing a dielectric liquid with constant velocity $\mathrm{v}$. Then -

Find the equivalent capacitance between $\mathrm{A}\mathrm{ }\&\mathrm{ }\mathrm{B}$ if each capacitor has capacitance C?