हिंदी
कर्नाटक बोर्ड पी.यू.सी.पीयूसी विज्ञान 2nd PUC Class 12

Determine the Capacitance of the Capacitor.

Advertisements
Advertisements

प्रश्न

A spherical capacitor has an inner sphere of radius 12 cm and an outer sphere of radius 13 cm. The outer sphere is earthed and the inner sphere is given a charge of 2.5 µC. The space between the concentric spheres is filled with a liquid of dielectric constant 32.

(a) Determine the capacitance of the capacitor.

(b) What is the potential of the inner sphere?

(c) Compare the capacitance of this capacitor with that of an isolated sphere of radius 12 cm. Explain why the latter is much smaller.

संख्यात्मक
Advertisements

उत्तर

Radius of the inner sphere, r2 = 12 cm = 0.12 m

Radius of the outer sphere, r1 = 13 cm = 0.13 m

Charge on the inner sphere, q = 2.5 µC = `2.5 xx 10^-6  "C"`

Dielectric constant of a liquid, ` in_"r" = 32`

(a) Capacitance of the capacitor is given by the relation

C = `(4piin_0in_1"r"_1"r"_2)/("r"_1 - "r"_2)`

Where,

`in_0` = Permittivity of free space = `8.85 xx 10^-12  "C"^2  "N"^-1  "m"^-2`

`1/(4piin_0) = 9 xx 10^9  "Nm"^2  "C"^-2`

∴ C = `(32 xx 0.12 xx 0.13)/(9 xx 10^9 xx (0.13 - 0.12))`

`≈ 5.5 xx 10^-9  "F"`

Hence, the capacitance of the capacitor is approximate `5.5 xx 10^-9  "F"`.

(b) Potential of the inner sphere is given by,

`"V" = "q"/"C"`

= `(2.5 xx 10^-6)/(5.5 xx 10^-9) = 4.5 xx 10^2  "V"`

Hence, the potential of the inner sphere is `4.5 xx 10^2  "V"`.

(c) Radius of an isolated sphere, r = 12 × 10−2 m

Capacitance of the sphere is given by the relation,

`"C'" = 4piin_0"r"`

= `4pi xx 8.85 xx 10^-12 xx 12 xx 10^-12`

= `1.33 xx 10^-11  "F"`

The capacitance of the isolated sphere is less in comparison to the concentric spheres. This is because the outer sphere of the concentric spheres is earthed. Hence, the potential difference is less and the capacitance is more than the isolated sphere.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 2: Electrostatic Potential and Capacitance - Exercise [पृष्ठ ९०]

APPEARS IN

एनसीईआरटी Physics Part I and II [English] Class 12
अध्याय 2 Electrostatic Potential and Capacitance
Exercise | Q 2.30 | पृष्ठ ९०

वीडियो ट्यूटोरियलVIEW ALL [2]

संबंधित प्रश्न

A capacitor of capacitance C is charged fully by connecting it to a battery of emf E. It is then disconnected from the battery. If the separation between the plates of the capacitor is now doubled, how will the following change?

(i) charge stored by the capacitor.

(ii) Field strength between the plates.

(iii) Energy stored by the capacitor.

Justify your answer in each case.


A capacitor of capacitance ‘C’ is charged to ‘V’ volts by a battery. After some time the battery is disconnected and the distance between the plates is doubled. Now a slab of dielectric constant, 1 < k < 2, is introduced to fill the space between the plates. How will the following be affected? (a) The electric field between the plates of the capacitor Justify your answer by writing the necessary expressions.


Define capacitor reactance. Write its S.I units.


Two metal spheres of capacitance C1 and C2 carry some charges. They are put in contact and then separated. The final charges Q1 and Q2 on them will satisfy


Each of the capacitors shown in figure has a capacitance of 2 µF. find the equivalent capacitance of the assembly between the points A and B. Suppose, a battery of emf 60 volts is connected between A and B. Find the potential difference appearing on the individual capacitors.


The separation between the plates of a parallel-plate capacitor is 0⋅500 cm and its plate area is 100 cm2. A 0⋅400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.


Suppose the space between the two inner shells is filled with a dielectric of dielectric constant K. Find the capacitance of the system between A and B.


The variation of inductive reactance (XL) of an inductor with the frequency (f) of the ac source of 100 V and variable frequency is shown in fig.

  1. Calculate the self-inductance of the inductor.
  2. When this inductor is used in series with a capacitor of unknown value and a resistor of 10 Ω at 300 s–1, maximum power dissipation occurs in the circuit. Calculate the capacitance of the capacitor.

Derive the expression for resultant capacitance, when the capacitor is connected in series.


During a thunder storm, the movement of water molecules within the clouds creates friction, partially causing the bottom part of the clouds to become negatively charged. This implies that the bottom of the cloud and the ground act as a parallel plate capacitor. If the electric field between the cloud and ground exceeds the dielectric breakdown of the air (3 × 106 Vm–1), lightning will occur.

  1. If the bottom part of the cloud is 1000 m above the ground, determine the electric potential difference that exists between the cloud and ground.
  2. In a typical lightning phenomenon, around 25 C of electrons are transferred from cloud to ground. How much electrostatic potential energy is transferred to the ground?

A capacitor is charged by a battery. The battery is removed and another identical uncharged capacitor is connected in parallel. The total electrostatic energy of resulting system ______.


To obtain 3 μF capacity from three capacitors of 2 μF each, they will be arranged ______.

The positive terminal of 12 V battery is connected to the ground. Then the negative terminal will be at ______.


A leaky parallel plate capacitor is filled completely with a material having dielectric constant K = 5 and electric conductivity σ = 7.4 × 10-12 Ω-1 m-1. If the charge on the plate at the instant t = 0 is q = 8.85 µC, then the leakage current at the instant t = 12 s is ______ × 10-1 µA.


Two identical capacitors are connected as shown and have an initial charge of Q0. The separation between the plates of each capacitor is d0. Suddenly the left plate of the upper capacitor and right plate of the lower capacitor start moving with speed v towards the left while the other plate of each capacitor remains fixed. `("given" (Q_0V)/(2d_0) = 10 A)`. The value of current in the circuit is ______ A.

 


Read the following paragraph and answer the questions.

A capacitor is a system of two conductors separated by an insulator. The two conductors have equal and opposite charges with a potential difference between them. The capacitance of a capacitor depends on the geometrical configuration (shape, size and separation) of the system and also on the nature of the insulator separating the two conductors. They are used to store charges. Like resistors, capacitors can be arranged in series or parallel or a combination of both to obtain the desired value of capacitance.
  1. Find the equivalent capacitance between points A and B in the given diagram.
  2. A dielectric slab is inserted between the plates of the parallel plate capacitor. The electric field between the plates decreases. Explain.
  3. A capacitor A of capacitance C, having charge Q is connected across another uncharged capacitor B of capacitance 2C. Find an expression for (a) the potential difference across the combination and (b) the charge lost by capacitor A.
    OR
    Two slabs of dielectric constants 2K and K fill the space between the plates of a parallel plate capacitor of plate area A and plate separation d as shown in the figure. Find an expression for the capacitance of the system.
     

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K1 and K2 with thickness `3/8  d and d/2`, respectively, are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K1 = 1.25 K, the value of K1 is:


Which expression gives the capacitance of a capacitor?


In a series combination, which quantity is the same on each capacitor?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×