WBCHSE Class 12 Physics Capacitance And Capacitor Short Answer Questions

WBCHSE Class 12 Physics Capacitance And Capacitor Short Question And Answers

Question 1. In a parallel plate capacitor with air between the plates, each plate has an area of 6 x 10^-3 m² and the distance between the plates is 3mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?
Answer:

α = 6 x 10^-3 m², d = 3 mm = 3 x 10^-3 m, V = 100 V

⇒ \(C_0=\frac{\epsilon_0 \alpha}{d}=\frac{\left(8.85 \times 10^{-12}\right)\left(6 \times 10^{-3}\right)}{3 \times 10^{-3}}\)

= 177 x 10^-11 F

q = 10 V

= 1.77 x 10^-11 x 100

= 1.77 x 10^-9 C

Question 2. Explain what would happen if the capacitor in Q.1, a 3 mm thick mica sheet (dielectric constant= 6) were inserted between the plates while the voltage supply remained connected.
Answer:

While the voltage supply remained connected, the voltage remained constant.

Capacity increases to

C’ = kC₀

= 6 x 1.77 x 10^-11F

= 1.062 x 10^-10 F

Charge increases to

q’ = C’V

= 1.062 x 10^-10 x 10²C

= 1.062 x 10^-8C

Question 3. A 600 pF capacitor is charged by a 200 V supply.It is then disconnected from the supply and is connected to another uncharged 600 pF capacitor. How much electrostatic energy is lost in the process?
Answer:

Here, C = 600 pF and V = 200 V.

∴ E = \(\frac{1}{2}\) CV²

= \(\frac{1}{2}\) x 600 x 10^-12 x (200)²

= 12 X 10^-6 J

When this capacitor is connected with another capacitor of the same capacity, the potential will be shared equally and the capacitance will be added.

∴ V’ = \(\frac{V}{2}\) 100 V and C’ = 2C = 1200 x 10^-12F

∴ \(E^{\prime}=\frac{1}{2} C^{\prime} V^{\prime 2}\)

= \(\frac{1}{2} \times 1200 \times 10^{-12} \times(100)^2\)

= 6 x 10^-6 J

Loss of electrostatic energy

= E- Ef

= (12-6) x 10^-6 J

= 6 x 10^-6 J

Question 4. An electric technician requires a capacitance of 2μF in a circuit across a potential difference of 1 kV. A large number of 1μF capacitors are available to him each of which can withstand a potential difference of not more than 400 V. Suggest a possible arrangement that requires the minimum number of capacitors.
Answer:

Let a combination of capacitors be arranged in which m rows with n capacitors each are connected in parallel.

∴ 400n = 1000 [∴ lkV = 1000 V]

or, n = \(\frac{1000}{400}=2.5 \approx 3\)

The equivalent capacity of each row

⇒ \(C=\frac{1}{\frac{1}{1}+\frac{1}{1}+\frac{1}{1}} \mu \mathrm{F}=\frac{1}{3} \mu \mathrm{F}\)

The equivalent capacitance of m such connected in parallel should be 2μF.

But C_eq= mC

∴ \(m=\frac{C_{\mathrm{eq}}}{C}=\frac{2}{1 / 3}=6\)

Total number of capacitors

= 3 x 6

= 18

They should be arranged in 6 rows having 3in each row and connected in parallel.

Question 5. The plates of a parallel plate capacitor have an area of 90 cm2 each and are separated by 2.5 mm. The capacitor is charged by connecting it to a 400 V supply.

1. How much electrostatic energy is stored by the capacitor?
2. View this energy as stored in the electrostatic field between the plates and obtain the energy per unit volume(u).
3. Arrive at a relation between u and the magnitude of the electric field between the two plates.

Answer:

Capacitance,

⇒ \(C=\frac{\epsilon_0 \alpha}{d}\)

= \(\frac{8.854 \times 10^{-12} 90 \times 10^{-4}}{2.5 \times 10^{-3}}\)

= \(3.187 \times 10^{-11} \mathrm{~F}\)

1. Stored energy = \(\frac{1}{2} C V^2=\frac{1}{2} \times 3.187 \times 10^{-11} \times 400^2\)

= 2.55 X 10^-6 J

2. Energy per unit volume,

⇒ \(u=\frac{2.55 \times 10^{-6}}{90 \times 10^{-4} \times 2.5 \times 10^{-3}}=0.113 \mathrm{~J} \cdot \mathrm{m}^{-3}\)

3. \(=\frac{\frac{1}{2} C V^2}{\alpha d}=\frac{1}{2} \frac{C}{\alpha d} \cdot(E d)^2\left[∵ E=\frac{V}{d}\right]\)

⇒ \(\frac{1}{2} \cdot \frac{\epsilon_0 \alpha}{d} \cdot \frac{E^2 d^2}{\alpha d}=\frac{1}{2} \epsilon_0 E^2\)

Question 6. Show that the force on each plate of a parallel plate capacitor has a magnitude equal to \(\frac{1}{2}\)QE, where Q is the charge on the capacitor and E is the magnitude of the electric field between the plates. Explain the origin of the factor\(\frac{1}{2}\).
Answer:

Let the application of force F increase the distance between the plates by Δx.

∴ Work done by the external force =F.Δx

This increases the energy.

∵ Energy density = u

So, increase in energy = u(αΔx) [α = area of the plate]

F- Δx = uα Δx

or, F = uα

∴ \(F=\frac{1}{2} \epsilon_0 E^2 \cdot \alpha \quad\left[∵ u=\frac{1}{2} \epsilon_0 E^2\right]\)

∴ \(F=\frac{1}{2} \epsilon_0 \alpha E \cdot E=\frac{1}{2} Q E\)

The factor \(\frac{1}{2}\) arises in this relation due to the fact that average \(\frac{E}{2}\) of 0 (field inside the conductor) and E (field outside the conductor) is taken.

Question 7.

1. What meaning would you give to the capacitance of a single conductor?

2. Guess a possible reason why water has a much greater dielectric constant (= 80) than say, mica (= 6).

Answer:

1. A single conductor may be considered as a capacitor whose second plate is situated at infinity.

2. Water molecules are polar molecules. Hence these molecules have their own dipole moments which contributes to the high value of its dielectric constant. But the molecules of mica are not polar

Question 8. In a Van de Graaff generator a spherical metal shellin to be a 15 x 10⁶ V electrode. The dielectric strength of the gas surrounding the electrode is 5 x 10⁷ V.m^-1. What is the minimum radius of the spherical shell required?
Answer:

Minimum radius of the shell,

⇒ \(=\frac{V}{E} \doteq \frac{15 \times 10^6}{5 \times 10^7}\)

= 3 x 10^-1 m

=30 cm

This result shows why an electrostatic generator which requires a small charge to acquire a high potential cannot be built with a very small shell.

Question 9. One evening a man fixes a two-metre high insulating slab carrying on its top a large aluminium sheet of area 1 m² outside his house. Will he get an electric shock if he touches the metal sheet the next morning?
Answer:

Yes, the earth, the aluminum sheet with the dielectric slab in between will form a capacitor. The potential of the aluminum sheet will rise due to the downpour of atmospheric charge during the night. When a person touches the sheet, the accumulated charge will flow through his body to Earth. This will constitute an electric current and the person will experience an electric shock.

Question 10. Two capacitors of capacitances 5μF and 10 juF are charged to 16 V and 10 V respectively. Find what will be the common potential when they are connected in parallel to each other.
Answer:

Initially, the charge on the first capacitor,

Q₁ = C₁V₁

= (5 x 10^-6) x 16

= 80 x 10^-6 C

and charge on the second capacitor,

Q₂ = C₂V₂

= (10 x 10^-6) x 10

= 100 x 10^-6 C

∴ Net charge, Q = Q₁ + Q₂

= 180 x 10^-6 C

Equivalent capacitance of the parallel combination,

C = C₁ + C₂

= (5 + 10)μF

= 15 x 10^-6 F

∴ The common potential of the combination,

⇒ \(V=\frac{Q}{C}=\frac{180 \times 10^{-6}}{15 \times 10^{-6}}=12 \mathrm{~V}\)

Question 11. What is understood by the capacitance of a capacitor? A 900 pF capacitor is charged to 100 V by a battery. How much energy is storedin the capacitor?
Answer:

C = 900 pF

=900 x 10^-12F

=9 X 10^-10F

∴ Stored energy \(\frac{1}{2} C V^2=\frac{1}{2} \times\left(9 \times 10^{-10}\right) \times 100^2\)

= 4.5 x 10^-6J

Question 12. A glass slab is introduced between the plates of a parallel plate capacitor. Does the capacitance of the capacitor increase, decrease, or remain unchanged? Two capacitors of capacitance 5μF and 10μF are charged to 16 V and 10 V respectively. Find the common potential when they are connected in parallel.
Answer:

Due to the insertion of a glass slab between the plates of a parallel plate capacitor, the capacitance will increase. If K is the dielectric constant of a glass slab, its capacitance will be K times its previous value.

Question 13. 64 identical water drops coalesce to form a larger drop. If the nature and amount of charge are the same for all the drops, calculate the potential, capacitance, and stored energy of the larger drop.
Answer:

Let the radii of small and large drops be r and R respectively and the charge of each small drop be Q.

According to the question,

⇒ \(\frac{4}{3} \pi R^3=64 \times \frac{4}{3} \pi r^3\)

or, R = 4r

Potential ofsmall drop, V = \(\frac{Q}{r}\)

or, Q = Vr

Potential of large drop, \(V^{\prime}=\frac{64 Q}{R}=\frac{64 V r}{4 r}\)

= 16V

i.e., the potential of a large drop will be 16 times the potential of a small drop.

As the radius of the large drop is 4r, its capacitance, C = 4r.

The energy storedin the large drop,

⇒ \(E=\frac{1}{2} C V^{\prime 2}=\frac{1}{2} \times 4 r \times(16 V)^2\)

= 512rV²

Question 14. Deduce the expression for the electrostatic energy storedin a capacitor of capacitance C and having charge Q. How will the electric field inside the capacitor be affected when it is completely filled with a dielectric material of dielectric constant K?
Answer:

When the capacitor is filled completely with a dielectric material of dielectric constant K, then capacitance becomes, C’ = kC.

∴ Changed potential difference,

⇒ \(V^{\prime}=\frac{Q}{C^{\prime}} \quad[… Q=\text { constant }]\)

⇒ \(\frac{Q}{K C}=\frac{V}{K}\)

∴ Changed electric field,

⇒ \(E^{\prime}=\frac{V^{\prime}}{d}=\frac{V}{\kappa d}=\frac{E}{K}\)

∴ The final electric field will be \(\frac{1}{k}\) times of its previous value.

Question 15. A slab of material of dielectric constant K has the same area as that of the plates of a parallel plate capacitor but has the thickness d/2, where d is the separation between the plates. Find out the expression for its capacitance when the slab is inserted between the plates of the capacitor
Answer:

Before insertion of the dielectric slab,

⇒ \(C_1=\frac{\epsilon_0 \alpha}{d}[\alpha=\text { area of each plate }]\)

After insertion of a dielectric slab of thickness t,

⇒ \(C_2=\frac{\epsilon_0 \alpha}{d-t\left(1-\frac{1}{k}\right)}\)

= \(\frac{\epsilon_0 \alpha}{d-\frac{d}{2}\left(1-\frac{1}{k}\right)}\left[∵ t=\frac{d}{2}\right]\)

⇒ \(=\frac{\epsilon_0 \alpha}{d\left(\frac{1}{2}+\frac{1}{k}\right)}\)

= \(\frac{C_1}{(\frac{1}{2}+\frac{1}{k})}[∵ C_1]\)

= \(\frac{\epsilon_0 \alpha}{d}\)

Question 16. A parallel plate capacitor of capacitance C is charged to a potential V. It is then connected to another uncharged capacitor having the same capacitance. Find out the ratio of the energy stored in the combined system to that stored initially in the single capacitor.
Answer:

Let the charge on the charged capacitor be Q.

∴ Energy stored,

⇒ \(U_1=\frac{Q^2}{2 C}\)

when another uncharged similar is connected with the first capacitor, the charge on the system remains

constant and the capacitance becomes C₁ = 2C.

∴ The energy stored in the system

⇒ \(U_2=\frac{Q^2}{4 C}\)

∴ \(U_2: U_1=\frac{Q^2}{4 C}: \frac{Q^2}{2 C}=1: 2\)

Question 17. Two capacitors of capacitance 10μF and 20μF are connected in series with a 6 V battery. After the capacitors are fully charged, a slab of dielectric constant (AT) is inserted between the plates of the two capacitors. How will the following be affected after the slab is introduced?

1. The potential difference between the plates of the capacitors
2. The charges on the two capacitors
3. The electric field energy storedin the capacitors

Answer:

Here, equivalent capacitance,

⇒ \(C_{\text {eq }}=\frac{C_1 C_2}{C_1+C_2}\)

= \(\frac{10 \times 10^{-6} \times 20 \times 10^{-6}}{(10+20) \times 10^{-6}}\)

= 6.67 x 10^-6F

Hence, the charge becomes

∴ Q = C_eqV

= 6.67 x 10^-6 x 6

= 4 x 10^-5C

Now, \(V_1=\frac{Q}{C_1}=\frac{4 \times 10^{-5}}{10 \times 10^{-6}}=4 \mathrm{~V}\)

⇒ \(V_2=\frac{Q}{C_2}=\frac{4 \times 10^{-5}}{10 \times 10^{-6}}=2 \mathrm{~V}\)

1. The potential difference between the plates of the capacitors would not change since the plates would attain the same potential as long as the battery is connected.

∴ V’₁ = 4V and V’₂ = 2V

2. After the dielectric slab is introduced, the capacitance becomes,

C’ = kC

Now, Q = CV

∴ Q’ = C’V = kCV = KQ

Hence, Q’₁ = QkC

= (4 x 10^-5)k C

Q’2 = (4 x 10^-5)kC

3. The electric field energy stored in the capacitors,

⇒ \(U=\frac{1}{2} C V^2\)

∴ \(U_1=\frac{1}{2} C_1 V_1^2=\frac{1}{2} \times\left(10 \times 10^{-6}\right) \times(4)^2\)

= 8 x 10^-5J

⇒ \(U_2=\frac{1}{2} C_2 V_2^2=\frac{1}{2} \times\left(20 \times 10^{-6}\right) \times(2)^2\)

= 4 x 10^-5J

After the introduction of the dielectric slab,

⇒ \(U_1^{\prime}=\kappa U_1=\left(8 \times 10^{-5}\right) \kappa \mathrm{J}

and U_2^{\prime}=\kappa U_2=\left(4 \times 10^{-5}\right) \kappa \mathrm{J}\)

Question 18. 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?

1. Charge stored by the capacitor
2. Field strength between the plates
3. Energy stored by the capacitor

Answer:

For the capacitor:

1. The charge stored on the capacitor does not change because of the law of conservation of charges.

2. The field strength between the plates is,

⇒ \(E=\frac{\sigma}{\epsilon_0}=\frac{Q}{A \epsilon_0}\)

Hence, we see that the field strength is independent of the distance between the plates. So, the field strength also remains the same.

3. The energy storedin the capacitor is,

⇒ \(U=\frac{Q^2}{2 C}\)

Now, when the distance between the plates is doubled, the capacitance becomes half. Hence, the energy stored will also double.

Question 19. 

1. Find the equivalent capacitance between A and the combination given below. Each capacitor is of 2μF capacitance.

2. If a source of 7 V is connected across AB, how much charge is drawn from the source and what is the energy storedin the network?

Answer:

1. In the circuit C₂, C₃ and C4 are in parallel combination.

C_p = C₂ + C₃ + C₄

= 2 + 2 + 2

= 6μF

∴ Equivalent capacitance between A and B,

⇒ \(\frac{1}{C_{\text {eq }}}=\frac{1}{C_1}+\frac{1}{C_P}+\frac{1}{C_5}=\frac{1}{2}+\frac{1}{6}+\frac{1}{2}=\frac{7}{6}\)

or, \(C_{\mathrm{eq}}=\frac{6}{7}=0.86 \mu \mathrm{F}\)

2. \(Q=C_{\mathrm{eq}} V=0.86 \times 7=6 \mu \mathrm{C}\)

Energy \(E=\frac{1}{2} Q V=\frac{1}{2} \times 6 \times 7=21 \mu \mathrm{J}\)

Question 20. A 12 pF capacitor is connected to a 50 V battery. How much electrostatic energy is stored in the capacitor? If another capacitor of 6 pF is connected in series with it with the same battery connected across the combination, find the charge stored and the potential difference across each capacitor.
Answer:

Electrostatic energy storedin the capacitor,

⇒ \(U=\frac{1}{2} C V^2=\frac{1}{2} \times 12 \times 10^{-12} \times(50)^2=1.5 \times 10^{-8} \mathrm{~J}\)

When 6 pF is connected in series with 12 pF, the charge is stored across each capacitor,

⇒ \(Q=\frac{C_1 C_2}{C_1+C_2} V=\frac{12 \times 6 \times 10^{-24}}{(12+6) \times 10^{-12}} \times 50=200 \mathrm{pC}\)

The potential difference across 12 pF is,

⇒ \(\frac{Q}{C_1}=\frac{200 \times 10^{-12}}{12 \times 10^{-12}}=16.67 \mathrm{~V}\)

The potential difference across 6 pF is,

⇒ \(\frac{Q}{C_2}=\frac{200 \times 10^{-12}}{6 \times 10^{-12}}=33.33 \mathrm{~V}\)

Question 21. Two identical capacitors of 12 pF each are connected in series across a battery of 50 V. How much electrostatic energy is stored in the combination? If these were connected in parallel across the same battery, how much energy would be stored in the combination now? Also, find the charge drawn from the battery in each case.
Answer:

For two identical capacitors connected in series, electrostatic energy is stored,

⇒ \(U_s=\frac{1}{2} C_s V^2\)

⇒ \(C_s=\frac{C_1 C_2}{C_1+C_2}=\frac{12 \times 12}{12+12}=6 \mathrm{pF} ; V=50 \mathrm{~V}\)

∴ \(U_s=\frac{1}{2} \times 6 \times 10^{-12} \times(50)^2=7.5 \mathrm{~nJ}\)

For two identical capacitors connected in parallel, electrostatic energy stored,

⇒ \(U_p=\frac{1}{2} C_p V^2\)

∴ \(U_p=\frac{1}{2} \times 24 \times 10^{-12} \times(50)^2=30 \mathrm{~nJ}\)

Charge drawn from the battery for a series combination of two identical capacitors,

Q_s = C_sV

= 6 x 10^-12 x 50

= 300 pc

Charge drawn from the battery for a parallel combination of two capacitors,

Q_p = C_pV

= 24 X 10^-12 x 50

= 1200 pC

Question 22. Two identical parallel plate capacitors A and B are connected to a battery of V volt with switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant K. Find the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric

 

Answer:

Initially, when the switch is closed, both the capacitors A and B are in parallel, and the energy storedin the capacitors is,

⇒ \(U_i=2 \times \frac{1}{2} C V^2=C V^2\)…(1)

When switch S is opened, B gets disconnected from the battery. Then capacitor B is isolated, and the charge on an isolated capacitor remains constant.

On the other hand, A remains connected to the battery and so the potential V remains constant.

When the capacitors are filled with dielectric, their capacitance increases to KC. Therefore, energy storedin B changes to \(\frac{Q^2}{2 \kappa C}\), where Q = CV is the charge on B. Energy storedin A changes to \(\frac{1}{2} \kappa C V^2\). Thus, the final total energy storedin the capacitor is,

⇒ \(U_f=\frac{1}{2} \frac{(C V)^2}{\kappa C}+\frac{1}{2} \kappa C V^2=\frac{1}{2} C V^2\left(\kappa+\frac{1}{k}\right)\)….(2)

From equations (1) and (2), we find

⇒ \(\frac{U_i}{U_f}=\frac{2 \kappa}{\kappa^2+1}\)