Sunday, January 31, 2021

H C Verma solutions, Capacitors, EXERCISES, Q1-Q10 Chapter-31, Concepts of Physics, Part-II

Capacitors


EXERCISES, Q1 - Q10


   1. When 1.0x10¹² electrons are transferred from one conductor to another, a potential difference of 10 V appears between the conductors. Calculate the capacitance of the two conductor system.  



Answer: The charge Q on the capacitor,

= 1x10¹²*1.602x10⁻¹⁹ C 

=1.602x10⁻⁷ C.

{Number of electrons multiplied by the charge on an electron} 

V = 10 V,

Hence the capacitance, C =Q/V

→C = 1.602x10⁻⁷/10 F

      =1.602x10⁻⁸ F.

      




 

    2. A The plates of a parallel plate conductor are made of circular discs of radii 5.0 cm each. If the separation between the plates is 1.0 mm, what is the capacitance? 



Answer: The radius of the discs, r =5.0 cm

=0.05 m  

 Area of the plate A =πr²

→A =π(0.05)² =7.85x10⁻³ m².

d = 1.0 mm

  = 1.0x10⁻³ m

Hence capacitance C = εₒA/d

   = 8.854x10⁻¹²*7.85x10⁻³/1x10⁻³ F

  = 69.5x10⁻¹² F

  = 6.95x10⁻¹¹ F

  = 6.95x10⁻⁵ µF.  







 

   3. Suppose one wishes to construct a 1.0-farad capacitor using circular discs. If the separation between the discs be kept 1.0 mm, what would be the radius of the discs?    



Answer: Given, C = 1.0 F,

d = 1.0 mm = 0.001 m 

Let the radius of the disc = r, 

Area of the circular disc, A =πr². 

 Capacitance, C = εₒA/d

→ 1.0 = 8.85x10⁻¹²*πr²/0.001

→πr² =1.0/8.85x10⁻⁹

→r² =36x10⁶ m²

→r =6x10³ m

→r = 6.0 km.      






 

   4. A parallel plate capacitor having a plate area of 25 cm²and separations of 1.00 mm is connected to a battery of 6.0 V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?   



Answer: A =25 cm² =2.5x10⁻³ m²,

d = 1.00 mm =0.001 m,

Capacitance, C = εₒA/d 

→C =8.85x10⁻¹²*2.5x10⁻³/0.001 

→C =2.21x10⁻¹¹ F


Also, the charge flowed through the battery, 

Q = CV

→Q =2.21x10⁻¹¹*6 C 

→Q =1.33x10⁻¹⁰ C.


Now the work-done by the battery, 

W = QV

   = 1.33x10⁻¹⁰*6 J

   = 8.0x10⁻¹⁰ J.


      




 

   5. A parallel plate capacitor has a plate area of 25.0 cm² and a separation of 2.00 mm between the plates. The capacitor is connected to a battery of 12.0 V. (a) Find the charge on the capacitor. (b) The plate separation is decreased to 1.00 mm. Find the extra charge given by the battery to the positive plate.   



Answer: A = 25 cm² =2.5x10⁻³ m².

d = 2 mm =0.002 m

Potential difference, V =12 volts

Capacitance, C =εₒA/d

→C =8.85x10⁻¹²*2.5x10⁻³/0.002 F

  =1.11x10⁻¹¹ F


(a) Charge on the capacitor

Q = CV

  =1.11x10⁻¹¹*12

  =1.33x10⁻¹⁰ C


(b) Extra charge given by the battery

When the separation is decreased,

d' = 1 mm = 0.001 m.

Now the capacitance

C' = εₒA/d

  =8.85x10⁻¹²*2.5x10⁻³/0.001 F

  =2.22x10⁻¹¹ F

Now charge on the capacitor

Q' = C'V

 =2.22x10⁻¹¹*12

 =2.66x10⁻¹⁰ C

Hence the extra charge given by the battery = Q' -Q

 =2.66x10⁻¹⁰ -1.33x10⁻¹⁰ C

 =1.33x10⁻¹⁰ C.

 





 

   6. Find the charges on the three capacitors connected to a battery as shown in figure (31-E1). Take C₁ = 2.0 µF, C₂ = 4.0 µF, C₃ = 6.0 µF and V = 12 volts.   
The figure for Q - 6



Answer: Given that, 

C₁ = 2.0 µF, C₂ = 4.0 µF, C₃ = 60 µF and V = 12 volts. 

     From the figure, it is clear that the potential difference across each capacitor is the same i.e. 12 volts.  

Hence charge on C₁ =C₁V

    =(2 µF)*(12 volts)

    =24 C.


Charge on C₂ = C₂V

     =(4 µF)*(12 volts)

     =48 C.


Charge on C₃ = C₃V

    =(6 µF)*(12 volts)

    =72 C. 






 

   7. Three capacitors having capacitances 20 µF, 30 µF, and 40 µF are connected in a series with a 12 V battery. Find the charge on each of the capacitors. How much work has been done by the battery in charging the capacitors?   



Answer: Let the equivalent capacitance of the series combination = C.

Hence, 1/C =1/20 +1/30 +1/40

→1/C =(6 +4 +3)/120

→C =120/13 µF 


Charge on this equivant capacitor,

 Q= CV =(120/13)*12 µC

            =110 µC.

In series combination, each capacitor has an equal charge for any value of capacitance and the charge on the equivalent capacitor is also the same. The charge on internal plates (not directly connected to the battery) does not come from the battery but due to the polarization of electrons on connected plates due to potential difference. Hence, the charge on each capacitor =110 µC.


The work done by the battery,

W =QV =(110x10⁻⁶ F)*(12 volts)

    =1.32x10⁻³ J.   







 

   8. Find the charge appearing on each of the three capacitors shown in figure (31-E2).   
The figure for Q - 8



Answer: The capacitors B and C are in a parallel arrangement. Hence the equivalent capacitance of B and C, 

C' =4 µF +4 µF =8 µF.

Now this equivalent capacitor and A are in series connection. Hence the equivalent capacitance of A and C', let it be C", is

1/C" =1/8 +1/8 =2/8 =1/4

→C" = 4 µF.

Now the charge supplied by the battery,

Q =C"V =4*12 µC =48 µC.

Since the charge supplied by the battery and charges appearing on each capacitor in series is the same, 48 µC of charge will appear on the equivalent capacitor C' and 48 µC of charge will appear on the capacitor A. 

     Since the capacitors B and C are similar and in parallel connection, the 48 µC of charge will get equally distributed between them. Hence the charge on each of the capacitors B and C =48/2 µC =24 µC.            









    9. Take C₁ = 4.0 µF and C₂ = 6.0 µF in figure (31-E3). Calculate the equivalent capacitance of the combination between the points indicated.  

The figure for Q - 9



Answer: (a) Let us name the points as below.

Digram for Q-9

   Between the points A and D/F, the capacitors C₁ and C₂ are connected in parallel. Hence the equivalent capacitance of these two capacitors, C =C₁ +C₂ = 4 +6 µF =10 µF.

        Similar is the case between D/F and B. Hence equivalent capacitance between these two points is also 10 µF. Now the whole combination is as if two capacitors 10 µF each are connected in series between the points A and B. Now the equivalent capacitance of the whole combination is given as,

1/C' = 1/10 +1/10 =2/10 =1/5

→C' = 5 µF.


(b) Let us see the figure carefully. The upper four capacitors and lower four capacitors are in a similar combination as in problem (a). It has been shown encircled in the diagram below. 

Diagram for Q-9(b)

      So we can replace them with equivalent capacitances C' and C' as shown in the above diagram (c). Since these two are connected in parallel, their equivalent capacitance C =C'+C'

→C = 2C' =2*5 µF =10 µF.  

 

  





 

   10. Find the charge supplied by the battery in the arrangement shown in figure (31-E4).   
The figure for Q - 10



Answer: Both the capacitors are connected to the battery in a similar way and the potential difference across both of them is 10 V. So they are connected in parallel. Equivalent capacitance of these two, C = 5.0 µF +6.0 µF =11.0 µF.

Hence the charge supplied by the battery, Q =C*V

→Q =11.0*10 µC =110 µC. 

---------------------------------------------------

Click here for all links → kktutor.blogspot.com 

===<<<O>>>===


My Channel on YouTube  →  SimplePhysics with KK


Links to the Chapters











EXERCISES - Q51 to Q55


CHAPTER- 27-Specific Heat Capacities of Gases

CHAPTER- 26-Laws of Thermodynamics


CHAPTER- 25-CALORIMETRY

Questions for Short Answer

OBJECTIVE-I

OBJECTIVE-II


EXERCISES - Q-11 to Q-18


CHAPTER- 24-Kinetic Theory of Gases







CHAPTER- 23 - Heat and Temperature






CHAPTER- 17 - Light Waves




CHAPTER- 14 - Fluid Mechanics



CHAPTER- 13 - Fluid Mechanics


CHAPTER- 12 - Simple Harmonic Motion








CHAPTER- 11 - Gravitation




CHAPTER- 10 - Rotational Mechanics




CHAPTER- 9 - Center of Mass, Linear Momentum, Collision


CHAPTER- 8 - Work and Energy

Click here for → Question for Short Answers

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → Exercises (1-10)

Click here for → Exercises (11-20)

Click here for → Exercises (21-30)

Click here for → Exercises (31-42)

Click here for → Exercise(43-54)

CHAPTER- 7 - Circular Motion

Click here for → Questions for Short Answer 

Click here for → OBJECTIVE-I

Click here for → OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → EXERCISES (11-20)

Click here for → EXERCISES (21-30)

CHAPTER- 6 - Friction

Click here for → Questions for Short Answer

Click here for → OBJECTIVE-I

Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

---------------------------------------------------------------------------------

CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

-------------------------------------------------------------------------------

CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


--------------------------------------------------------------------------------------------------------------

CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

Posted by at No comments:
Labels: , , ,
Subscribe to: Posts (Atom)