# Breaking a Capacitor into Combinations

## Trending Questions

**Q.**

A parallel plate
capacitor with air between the plates has a capacitance of 8 pF (1pF
= 10^{−12} F). What will be the capacitance if the
distance between the plates is reduced by half, and the space between
them is filled with a substance of dielectric constant 6?

**Q.**A parallel plate capacitor has two layers of dielectric as shown in figure. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is:

- 32
- 43
- 12
- 13

**Q.**

A parallel plate capacitor with air as medium between the plates has a capacitance of 10μF. The area of capacitor is divided into two equal halves

and filled with two media as shown in the figure having dielectric constant k_{1}=2 and k_{2}=_{4}. The capacitance of the system will now be

30μF

40μF

10μF

20μF

**Q.**A wire of magnetic dipole moment M and L is bent into shape of a semicircle of raduis r. What will be its new dipole moment?

**Q.**The area of the plates of a parallel plate condenser is A and the distance between the plates is 10mm. There are two dielectric sheets in it, one of the dielectric constant is 10 and thickness 6mm and the other of dielectric constant 5 and thickness 4mm. The capacity of the condenser is: ______.

**Q.**Separation between the plates of a parallel plate capacitor is and the area of each plate is A. When a slab of material of dielectric constant and thickness t(t<d) is introduced between the plates, its capacitance becomes

**Q.**

When a dielectric slab of thickness 6cm is introduced between the plates of parallel plate condenser, it is found that the distance between the plates has to be increased by 4cm to restore the capacity to original value. The dielectric constant of the slab is

- 1.5
- 2/3
- 4
- 3

**Q.**does capacitance depends upon the distance between the two plate of capacitors as depicted by formula?

**Q.**Capacitance of a capacitor becomes 43 times its original value if a dielectric slab of thickness t=d2 is inserted between the plates (d = separation between the plates). Then the dielectric constant of the slab is

**Q.**A parallel-plate capacitor having plate area A and plate separation d is joined to a battery of emf E and internal resistance R at t=0. Consider a plane surface of area A/2, parallel to the plates and situated symmetrically between them. If the displacement current through this surface as a function of time is given as:

id=EnRe−tdε0AR

Where, ε0 is the permittivity of free space.

The value of n is :

**Q.**A slab of dielectric constant K has the same crosssectional area as the plates of a parallel plate capacitor and thickness 34d, where d is thenseparation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:

(Given C0=capacitance of capacitor with air as medium between plates.)

**Q.**A parallel plate condenser initially has air as a medium between the plates. If a slab of dielectric constant 5 having thickness half the distance of separation between the plates is introduced, the percentage increase in its capacity will be

- 66.7%
- 50%
- 75%
- 33.3%

**Q.**When the switch is closed, then the initial current through 1 Ω resistor in the given circuit is (Consider the capacitor uncharged initially)

- 12 A
- 4 A
- 107 A
- 3 A

**Q.**30. A capacitor as shown in figure has square plates length l each and are inlclined at angle theta with one another.for small value of theta, capacitance is.given by.

**Q.**

Find equivalent capacitance of system if plate area is A.

[Dielectric slab of constant 2K and K are of same volume and half of the volume of the slab of constant 3K.]

- 5 ϵ0AKd
- 136ϵ0AKd
- 6 ϵ0AKd
- 157ϵ0AKd

**Q.**The plates of a parallel plate capacitor are charged up to 100 V. Now, after removing the battery, a 2 mm thick plate is inserted between the plates. Then, to maintain the same potential difference, the distance between the plates is increased by 1.6 mm. Dielectric constant of the plate is

- 4
- 2.5
- 5
- 1.25

**Q.**A 10μF capacitor and a 20μF capacitor are connected in series across a 200 V supply line. The charged capacitors are then disconnected from the line and reconnected with their positive plates together and negative plates together and no external voltage is applied. What is the potential difference across each capacitor?

**Q.**

A parallel plate capacitor with air between the plates has a capacitance of $8pF(1pF={10}^{-12}F)$. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6$?

**Q.**Figure shows five capacitors connected across a 12 V power supply. What is the charge on the 2 μF capacitor ?

- 6 μC
- 8 μC
- 10 μC
- 12 μC

**Q.**

A parallel plate air condenser has a capacity of 4 μF. It is filled with two different dielectrics of constants 4 and 6 and they fill the upper and lower halves of the space as shown. The capacity now will be

20 μF

- 19.2 μF
9.6 μF

10 μF

**Q.**The thickness of earlier between the two coating of spherical conductor is 2 cm the capacitor has the same capacitance is a square of 1.2 metre diameter find the radii of the two sphere

**Q.**A capacitor of 4μF charged at 50V is connected with another capacitor of 2μF charged at 100V, in such a way that plates with similar charges are connected together. Before joining and after joining, the total energy will be

- 1.5×10−2J & 1.33×10−2 J
- 1.33×10−2J & 1.5×10−2 J
- 1.5×10−2J & 2.67×10−2 J
- 2.67×10−2J & 1.5×10−2 J

**Q.**A parallel-plate capacitor is constructed using three different dielectric materials as shown in the figure. What is the capacitance across P and Q?

- (K12+K2K3K2+K3)ε0At
- (K1+K2K3K2+K3)ε0At
- (K1+2K2K3K2+K3)ε0At
- (K1+K2K32(K2+K3))ε0At

**Q.**The capacitance of a parallel plate capacitor having plate area A and separation 2d with dielectrics inserted as shown in figure is

- 3kϵ0A2d
- 5kϵ0A12d
- kϵ0Ad
- 7kϵ0A12d

**Q.**142.a parallel plate air capacitor has a of 1 mue F . the surface area of each plate A , their separation is d . a sheet of dilectric constant 4, surface area A and tthicness d/2 is introduced completely in the intervening space of the capacitor .what will be its capacitance now?

**Q.**78. In the fig given find effective capacitance between X and Y if each capacitor is 1 micro farad

**Q.**Name the component X used to get steady DC output from the pulsating voltage.

- Resistor
- None of these
- Inductor
- Capacitor

**Q.**A parallel plate capacitor with plate area 100 cm2 and seperation between the plates 1.0 cm is connected across a battery of emf 24 V. The force of attraction between the plates is

- 1.6×10−5 cm
- 1.0×10−7 N
- 4×10−5 N
- 2.5×10−7 N

**Q.**An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant. If during this process the relation of pressure P and volume V is given by PVn=constant, then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively).

- n=CPCV
- n=C−CPC−CV
- n=CP−CC−CV
- n=C−CVC−CP

**Q.**Three plates P, Q, R each of area A have separation d between P and Q and also between Q and R. The energy stored when the plates are fully charged is:-

- 2ε0AV2d
- ε0AV2d
- ε0AV22d
- ε0AV24d