H C VERMA Solutions for Class 12-science Physics Chapter 9 - Capacitors

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Chapter 9 - Capacitors Exercise 165

Question 1

When 1.0×1012.electrons are transferred from one conductor to another, a potential difference of 10V appears between the conductors. Calculate the capacitance of the two-conductor system.

Solution 1

  

  

  

  

Question 2

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

Solution 2

  

  

C=6.95  

Question 3

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

Solution 3

  

1=  

r=6000m

r=6km

Question 4

A parallel-plate capacitor having plate area 25cm2 and separation 1.00mm is connected to a battery of 6.0V. Calculate the charge flown through the battery. How much work has been done by the battery during the process?

Solution 4

Capacitance

  

=  

=2.2×10-11

Charge flown Q=CV

=  

Q=1.33  

Work done by battery

W=QV

=1.33  

W=8  

Question 5

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

Solution 5

C=  

C=  

C=11 F

(a) Charge

Q=CV

Q=  

Q=1.33  

(b)New capacitance

 =  

=  

 =22.12 F

New charge on capacitor

 = V

= (12)

 =2.65 C

Extra charge flown = -Q

=  

 C

Question 6

Find the charges on the three capacitors connected to a battery as shown in figure.

  

 

 

Solution 6

Potential difference across each capacitor is V as all are in parallel combination.

Charge on capacitor  = V =  =24  

Charge on capacitor  = V =  =48  

Charge on capacitor  = V= =72  

Question 7

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

Solution 7

Equivalent capacitance

  = + +  

  = + +  

 =9.2  

Charge on each capacitor will be same as all are in series combination

Q =   

Q=9.2×12

Q = 110μC

Work done by the battery = QV

=(110)(12)

=1.33×10-3 J

Question 8

Find the charge appearing on each of the three capacitors shown in figure

 

 

Solution 8

Equivalent capacitance

 =  +  

 = +  

 4μF

Charge flown by battery =   

=(4μF)(12)

=48μC

Charge on capacitor A = 48μC

Charge on capacitor B =   = 24μC

Charge on capacitor C =   = 24μC

Question 9

Take C1=4.0 F and C2=6.0μF in figure. Calculate the equivalent capacitance of the combination between the points indicated.

 

 

Solution 9

(a)

  

 

  

  

(b)

  

 

  

  

Chapter 9 - Capacitors Exercise 166

Question 10

Find the charge supplied by the battery in the arrangement shown in figure

 

 

Solution 10

 

  

Charge flown Q=CV

=11×10

  

Question 11

The outer cylinders of two cylindrical capacitors of capacitance 2.2μF each, are kept in contact and the inner cylinders are connected through a wire. A battery of emf 10 V connected as shown in figure. Find the total charge supplied by the battery to the inner cylinders.

 

 

Solution 11

Both cylindrical capacitor are in parallel combination as potential difference across them is some

 +   

= 2.2+2.2

 4.4μF

Total charge supplied Q=CeqV

=4.4×10

Q=44μC 

Question 12

Two conducting spheres of radii R1and R2 are kept widely separated from each other. What are their individual capacitance? If the spheres are connected by a metal wire, what will be the capacitance of the combination? Think in terms of series - parallel connections.

Solution 12

let change Q is given to the individual conducting sphere potential, V =   

V =   

C=  = 4 R

For radius  ,  = 4  

For radius   ,  = 4  

When connected by wire, there potential will become some.

Potential difference of conducting spheres with respect to infinity will be same. So, both are in parallel combination.

 +   

  4  

Question 13

Each of the capacitors shown in figure has a capacitance of 2μF. Find the equivalent capacitance of the assembly between the points A and B. Find the potential difference appearing on the individual capacitors.

 

 

Solution 13

For each row, equivalent capacitance is

  = + +  

 F

Now , all 3 rows are in parallel combination

 =  + +  = 2μF

Potential difference across each row is 60V and drop on each capacitor in row will be same as al have same capacitance.

i.e. potential drop = 20V on each.

Question 14

It is required to construct a 10μF capacitor which can be connected across a 200V battery. Capacitors of capacitance 10μF are available but they can withstand only 50V. Design a combination which can yield the desired result.

Solution 14

Voltage across each row =200V

Let x capacitors are used in each row with breaking potential of 50V

So ,

50(x)=200

X = 4

Effective capacitance of row =  F

Now, let y rows be connected in

 = = 10μF

Y=4

So ,

Combination of 4 rows each is having 4 capacitors.

Question 15

Take the potential of the point B in figure to be zero. (a) Find the potentials at the points C and D. (b) If a capacitor is connected between C and D, what charge will appear on this capacitor?

 

 

Solution 15

(a) Potential drop across 4 μF and μF is 50V and both are in series combination so charge will be same.

So, potential drop on each capacitor will be in inverse ratio of capacitance

  =  =  

Potential drop across 8 F =  

 -  =   

 volt

Similarly

  =  =  

Potential drop across 6μF =   

 -  = volt

 =  volt

b. no change will flow as   

Question 16

Find the equivalent capacitance of the system shown in figure between the points a and b.

 

 

Solution 16

capacitors  and   are in series and then will be in parallel with   

  + +   

 =   +  

Question 17

A Capacitor is made of a flat plate of area A and a second plate having a stair-like structure as shown in figure. The width of each stair is a and the height is b. Find the capacitance of the assembly.

 

  

 

Solution 17

Area of each stair facing the flat plate is same i.e.  

 

 

All capacitors are in parallel combination

  

  

  

Question 18

A cylindrical capacitor is constructed using two coaxial cylinders of the same length 10cm and of radii 2mm and 4mm. (a) Calculate the Capacitance. (b) Another capacitor of the same length is constructed with cylinders of radii 4mm and 8mm. Calculate the capacitance.

Solution 18

(a) capacitance

  

  

C = 8pF

(b) Same as ratio of radii and length of cylinders are same,

C = 8pF

Question 19

A 100pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20pF. What will be the new potential difference across the 100pF capacitor?

Solution 19

  

  

=20V

Question 20

Each capacitor shown in figure has a capacitance of 5.0μF. The emf of the battery is 50V. How much charge will flow through AB if the switch S is closed?

 

 

Solution 20

Initially ,

 

 

  Charge supplied by battery Q= CV

Q=  =  C

Now , when switch is closed , the capacitor gets short - circuited

  = 5+5 =10 F

Charge supplied by battery  V

= 10×50 = 500μC

Hence ,

Charge supplied =  -Q

= = 3.3 C

Question 21

The particle P shown in figure has a mass of 10mg and a charge of -0.01μC. Each plate has a surface area 100 cm2. On one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?

 

 

Solution 21

Potential drop on each capacitor   

Particle is in equilibrium then

Mg = qE

Mg = q  

  = V   

V=   

V = 43 m volt.

Question 22

Both the capacitors shown in figure are made of square plates of edges a. The separations between the plates of the capacitors are d1 and d2 as shown in the figure. A potential difference V is applied between the point a and b. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electron be projected so that it does not collide with any plate? Consider only the electric forces.

 

 

Solution 22

 

Potential difference across   will be inverse ratio as both are in series combination

  

  upper capacitor

  =  V

Electric field in upper capacitor

  =  =  

For electron

x-axis

y-axis

  = u

  = 0

  = 0

 =   

  =a

  

  =  t +   

  =  

  

t=   

  u  

u = a  

u =   

 

Chapter 9 - Capacitors Exercise 167

Question 23

The plates of a capacitor are 2.00cm apart. An electron-proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At the distance from the negative plate was the pair released?

Solution 23

 

For electron,

  

  

  

  

 --------(1)

For proton,

  

  

  

  

 --------(2)

Divide (1) and (2)

  

  

Question 24

Convince yourself that parts (a), (b), (c) of figure are identical. Find the capacitance between the points A and B of the assembly.

 

  

 

Solution 24

All the circuits are in balanced wheat stone symmetry so no current flows is 5μF capacitor

  

Question 25

Find the potential difference Va-Vb between the points a and b shown in each part of the figure

 

  

 

Solution 25

  

 

Nodal analysis at X

  

 --------------(1)

Nodal analysis at y

  

 ---------------(2)

Putting in (1)

  

  

So,

  

  

(b)

 

 

 

  

  

  

  

 

So,

  

(c)

 

 

Nodal Analysis at X

  

  

  

  

So,

  

(d)

 

Nodal analysis at Vb

  

  

  

So,

  

  

Question 26

Find the equivalent capacitance of the combination of the combination shown in figure between the indicated points.

 

  

 

Solution 26

(a) By input-output symmetry, Potential difference across 1 F must be same

 

Nodal Analysis at X

  

  

  

For charge

  

  

For charge

  

  

Total charge flown=  

  

  

  

(b)

 

 

By input - output symmetry,

Potential difference across 1μF, will be same.

Potential difference across 2μF, will be same and is equal to   .

Nodal analysis at x

  

  

Current in 1μF

  

Current in 2μF

  

Current in 3μF

  

Current flown in battery =

  

  

  

(c)

 

 

It is in wheat stone symmetry

So, 5  Capacitor is removed

  

  

  

(d) It is in wheat stone symmetry

So, all 6  capacitors will be removed

  

  

Question 27

Find the capacitance of the combination shown in figure between A and B

 

 

Solution 27

 

C1 and C2 are in series =  

This is in parallel with C3= 1+1=2  

This is in series with C4=  

This is in parallel with C5= 1+1=2  

This is in series with C6=  

This is in parallel with C7= 1+1=2  

This is in series with C8=  

  

Question 28

Find the equivalent capacitance of the infinite ladder shown in figure between the points A and B.

 

  

Solution 28

 

Let CAB=x then CCD=x

  

  

  

  

Therefore,

  

Question 29

A finite ladder is constructed by connecting several sections of 2μF, 4μF capacitor combination as shown in figure. It is terminated by a capacitor of capacitance C. What value should be chosen for C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between?

 

Solution 29

If capacitance of encircled part again comes C then it becomes independent of sections

  

  

  

  

  

(Not Possible)

Question 30

A charge of +2.0×10-8C is placed on the positive plate and a charge of -1.0×10-8C on the negative plate of a parallel-plate capacitor of capacitance 1.2×10-3μF.

Solution 30

 

Charge on the outer faces of the plates is average of all charges   

=  

 = 0.5 C

  = 2  - 0.5×10-8 

= 1.5  

Q=CV

1.5 V

V = 12.5 volt

Chapter 9 - Capacitors Exercise 168

Question 31

A charge of 20 micro coloumb is placed on the positive plate of an isolated parallel-plate capacitor of capacitance 10 microfarad. Calculate the potential difference developed between the plates.

Solution 31

Charge on each surface will be   = 10μC to have zero

Electric field inside plate

Effective charge on the capacitor = 10μC

Q=CV

10μ = 10μV

V = 1volt

Question 32

A charge of 1 μC is given to one plate of a parallel-plate capacitor of capacitance 0.1 μF and a charge of 2 μC is given to the other plate. Find the potential difference developed between the plates.

Solution 32

Outer plate will have average of charge

  

  

  = 1μC -   

=1-1.5

  = -0.5 μC

Effective charge on the capacitor = -0.5μC

Q = CV

0.5=0.1(v)

V = 5 volt

Question 33

Each of the plates shown in figure has surface area   on one side and the separation between the consecutive plates is 4.0 mm. The emf of the battery connected is 10 volts. Find the magnitude of the charge supplied by the battery to each of the plates connected to it.

 

  

 

Solution 33

 

Capacitance between each plates is same

C= =   

C = 24 F

Nodal analytic at X

  = 0

3X =20

X =   

Potential difference on each side of 10 V plate = 10 - X

=10 -   

=   volt

Charge on each side = CV

=  

=8 C

Total charge supplied by battery = 2  

 C

=0.16μC

Question 34

The capacitance between the adjacent plates shown in figure is 50 nF. A charge of l.0μC is placed on the middle plate, (a) What will be the charge on the outer surface of the upper plate? (b) Find the potential difference developed between the upper and the middle plates.

 

  

Solution 34

 

(a) To have zero electric field inside middle plate (1μC) is equally distributed on plate.

(b)   =0.5μC

Q = CV

0.5  -50 V

V=10volt

Question 35

Consider the situation of the previous problem. If l.O μC is placed on the upper plate instead of the middle, what will be the potential difference between (a) the upper and the middle plates and (b) the middle and the lower plates?

Solution 35

 

(a)   

Q = CV

0.5   = 50  

V= 10volt

(b)  C

Q = CV

0.5   = 50  

V=10volt 

Question 36

Two capacitors of capacitances 20.0 pF and 50.0 pF are connected in series with a 6.00 V battery. Find (a) the potential difference across each capacitor and (b) the energy stored in each capacitor.

Solution 36

(a)   =   (series combination )

So,   =   =   

Potential difference across 20pf =   = 4.29V

Potential difference across 50pf = 6-4.29 = 1.71 V

(b) energy stored in 20pf =  C  

=  = 184pJ

And in 50pF =   = 73.5 pJ

Question 37

Two capacitors of capacitances 4.0 μ F and 6.0 μF are connected in series with a battery of 20 V. Find the energy supplied by the battery.

Solution 37

  =   = 2.4μF

Energy supplied by battery = Q×V

=C  

=   

=960μJ

Question 38

Each capacitor in figure has a capacitance of 10μF. The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.

 

 

Solution 38

B and c are in parallel combination

 =  

=10+10

 20μF

All are in series combination , is charge will be same on each

Q=CV

  

  

  =   = 40V

  =   = 20V

  =  = 40V

Energy stored

  =   =  C  =   = 8mJ

  =  = 2mJ

Question 39

A capacitor with stored energy 4.0 J is connected with an identical capacitor with no electric field in between. Find the total energy stored in the two capacitors.

Solution 39

Initially , the potential on capacitor be V

  = C  4 J

Now ,

Energy loss   

  = 2J

Energy stored in two capacitors = 4J - 2J

=2J

Question 40

A capacitor of capacitance 2 0 μF is charged to a potential difference of 12 V. It is then connected to an uncharged capacitor of capacitance 4'0 μF as shown in figure (31-E22). Find (a) the charge on each of the two capacitors after the connection, (b) the electrostatic energy stored in each of the two capacitors and (c) the heat produced during the charge transfer from one capacitor to the other.

 

 

Solution 40

  

=  

  = 4V

(a) Charge on 2  =   = 8   

And 4  =   = 16  

(b) Energy stored = C  

In  =  = 16  

And In  =  = 32  

(c)   

=   

=96  

Question 41

A point charge Q is placed at the origin. Find the electrostatic energy stored outside the sphere of radius R centred at the origin.

Solution 41

Capacitance of sphere = 4  

Energy stored =  C  

=  

  

Question 42

A metal sphere of radius R is charged to a potential V. (a) Find the electrostatic energy stored in the electric field within a concentric sphere of radius 2 R. (b) Show that the electrostatic field energy stored outside the sphere of radius 2 R equals that stored within it.

Solution 42

  =   

Energy stored in spherical shell f width dr at distance r

dU =  . dr

U =   =   

U =   

From R to 2R

  =   

=   

  =  R  

  from 2R   

  

 R  

 . 

Question 43

A large conducting plane has a surface charge density   Find the electrostatic energy stored in a cubical volume of edge 1.O cm in front of the plane.

Solution 43

  =   

  =   =   

  =   

Energy stored = 5.6   J

Question 44

A parallel-plate capacitor having plate area 20   and separation between the plates l.00 mm is connected to a battery of 12.0 V. The plates are pulled apart to increase the separation to 2.0 mm. (a) Calculate the charge flown through the circuit during the process, (b) How much energy is absorbed by the battery during the process? (c) Calculate the stored energy in the electric field before and after the process, (d) Using the expression for the force between the plates, find the work done by the person pulling the plates apart, (e) Show and justify that no heat is produced during this transfer of charge as the separation is increased.

Solution 44

Initial capacitance   =   

=   

  =1.77 F

  

  =   

=   

  =0.88 F

  charge flown =  

 ( V

 (0.88 (12)

 C

  energy absorbed = (Q)(V)

=  

 J

  =   =   

= J

  =   =   

 6.35 J

Work done = force × displacement

W =   =   

W = 6.35 J

Work done   

So, no heat is parallel during transfer

Question 45

A capacitor having a capacitance of 100 μF is charged to a potential difference of 24 V. The charging battery is disconnected and the capacitor is connected to another battery of emf 12 V with the positive plate of the capacitor joined with the positive terminal of the battery, (a) Find the charges on the capacitor before and after the reconnection. (b) Find the charge flown through the 12 V batteries, (c) Is work done by the battery or is it done on the battery? Find its magnitude, (d) Find the decrease in electrostatic field energy, (e) Find the heat developed during the flow of charge after reconnection.

Solution 45

(a)   = C  

= (100)(24)

  =2400  

  = C  

 (100)(12)

  =1200  

(b) charge flown =   

= 2400-1200

= 1200  

(c) work is done on battery

W = charge × potential

= (1200μ)(12)

W = 14.4 mJ

(d)   =   

  =   

  

= 21600  

  21.6mJ

  + heat loss

  = 14.4 + heat loss

heat loss = 7.2mJ

Question 46

Consider the situation shown in figure. The switch S is open for a long time and then closed, (a) Find the charge flown through the battery when the switch S is closed, (b) Find the work done by the battery.

 

  

 

Solution 46

 

(a) initially   =   =   

And charge flown =   

Now, switch is closed

Charge flown =   

(b) work done = charge × potential

=   

=   

  =  =   

  + C   =   

  

  

  

(b) heat = work done - U

  

Heat =   

Chapter 9 - Capacitors Exercise 169

Question 47

A capacitor of capacitance 5.00 μF is charged to 24.0 V and another capacitor of capacitance 6'0 μF is charged to 12.0 V. (a) Find the energy stored in each capacitor, (b) The positive plate of the first capacitor is now connected to the negative plate of the second and vice versa. Find the. new charges on the capacitors, (c) Find the loss of electrostatic energy during the process, (d) Where does this energy go ?

Solution 47

(a) Energy stored   

In 5  =  = 1.44 mJ

In 6  =  = 0.432 mJ

(b)   

  

  

 on 5  C  = 5   = 21.8  

 on 6  C  = 6   = 26.2  

(c)   

  

 (d) This energy is dissipated as heat.

Question 48

A 5.0 μF capacitor is charged to 12 V. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires."

Solution 48

  

 

  = CV

  

  = 60μC

 

  

 

  = CV

  

  = 60μC

Now, charge flown through battery = 120μC

 + heat produced = workdone

 = 120×12

heat=1.44mJ

Question 49

The two square faces of a rectangular dielectric slab (dielectric constant 4.0) of dimensions 20 cm × 20 cm × l.O mm are metal-coated. Find the capacitance between the coated surfaces.

Solution 49

  

  

C= 1.42nF

Question 50

If the above capacitor is connected across a 6.0 V battery, find (a) the charge supplied by the battery, (b) the induced charge on the dielectric and (c) the net charge appearing on one of the coated surfaces.

Solution 50

(a) charge Q = CV

=   

Q = 8.5nc

(b) induced charge on dielectric

  = Q  

  

  = 6.4nc

  = Q -   

  - 6.4

 .1nc

Question 51

The separation between the plates of a parallel-plate capacitor is 0.500 cm and its plate area is 100 cm-square. 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.

Solution 51

capacitance

  

K = ( metal )

  

C = 88pF

Question 52

A capacitor stores 50 μC charges when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100 μC flows through the battery. Find the dielectric constant of the material inserted.

Solution 52

  = c;   = V

  

  cv (i)

  = c;   = V

  

  = CKV

 CKV (ii)

divided (ii) by (i)

k=3

Question 53

A parallel-plate capacitor of capacitance 5 μF is connected to a battery of emf 6 V. The separation between the plates is 2 mm. (a) Find the charge on the positive plate; (b) Find the electric field between the plates, (c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination, (d) How much charge has flown through the battery after the slab is inserted?

Solution 53

(a) charge Q = VC

=(5)(6)

  = 30  

(b) electric field =   

=   = 300  

(c) new capacitance   =   

Divide   =   

  =   

  = 8.33μF

(d) final charge on capacitor   = V

=   

  = 50μC

So, charge flown =   =   

=   

=20  

Question 54

A parallel-plate capacitor has plate area 100 cm-square and piate separation l.O cm. A glass plate (dielectric constant 6.0) of thickness 6.0 mm and an ebonite plate (dielectric constant 4.0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

Solution 54

capacitance =   

  

=44.25pF

Question 55

A parallel-plate capacitor having plate area 400 cm-square and separation between the plates l.O mm is connected to a power supply of 100 V. A dielectric slab of thickness 0.5 mm and dielectric constant 5.0 is inserted into the gap. (a) Find the increase in electrostatic energy, (b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy, (c) Why does the energy increase in inserting the slab as well as in taking it out?

Solution 55

initial capacitance

  =   =   

  = 3.54  F

  

 =   =   

 =   F

(a) change in energy stored

  -   

  

 1.18  

(b) charge on capacitor when dielectric is inside

  

=  

Q =   C

Now , battery is disconnected , then charge will remain constant energy stored

U =   

  =   

  =   

U =   

U = 1.92   

(c) during insertion capacitance increases , more charge flows from battery and hence battery supplies energy

During removal, energy increases, work has to be external agent to remove it.

Question 56

Find the capacitances of the capacitors shown in figure. The plate area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.

 

Solution 56

(a) The two parts of the capacitor are in series with capacitance C1 and C2

  

  

As they are connected in series the net capacitance would be

  

On substituting the value we get

  

(b) Here the Capacitor has three parts and are connected in series

  

  

  

  

On substituting values

  

(c)

  

  

These two are in parallel

C = C1+C2

  

Question 57

A capacitor is formed by two square metal-plates of edge a, separated by a distance d. Dielectrics of dielectric constants K1 and K2 are filled in the gap as shown in figure . Find the capacitance.

  

Solution 57

 

Elemental capacitor of width dx is assumed at distance from left end

d  =   ; d  =   

Both are in series , so equivalent capacitance is given by

  

  

  

C =   

  

=  ln  

C =  ln  =   

C =   

Question 58

Figure shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.

 

  

 

Solution 58

Initially

  =   ;   =   

And energy stored   =  C  +  C  = C  

Now ,

  = 3c

Charge and potential will remain constant for   and   respectively

  =   +   

  +   

  =   

So,

  

Question 59

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.

Solution 59

Initially charge  CV

Energy stored initially  C  

  

Now, when battery is disconnected charge remains constant.   

  = KC

  

  

  

Work done = change in energy

=   

W =   

Question 60

A capacitor having a capacitance of 100 μF is charged to a potential difference of 50 V. (a) What is the magnitude of the charge on each plate? (b) The charging battery is disconnected and a dielectric of dielectric constant 2.5 is inserted. Calculate the new potential difference between the plates, (c) What charge would have produced this potential .difference in absence of the dielectric slab, (d) Find the charge induced at a surface of the dielectric slab.

Solution 60

(a) charge Q = CV

  

Q = 5mc

(b) charge remains constant   = Q

  

  

 =20V

(c) charge = CV

  

(d) qin = q   

=5  

qin =3mc

Question 61

A spherical capacitor is made of two conducting spherical shells of radii a and b. The space between the shells is filled with a dielectric of dielectric constant K up to a radius c as shown in figure Calculate the capacitance.

 

 

Solution 61

One capacitor made by shells a and c and other capacitor by the shells b and c

  =   ;   =   

  

  

  

  

Chapter 9 - Capacitors Exercise 170

Question 62

Consider an assembly of three conducting concentric spherical shells of radii a, b and c as shown in figure, Find the capacitance of the assembly between the points A and B.

 

 

Solution 62

  

 

  

  

  

Question 63

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

Solution 63

  

  

  

  

Question 64

An air filled parallel-plate capacitor is to be constructed which can store 12μC of charge when operated at 1200V. What can be the minimum plate area of the capacitor? The dielectric strength of air 3×106V/m.

Solution 64

  

  

  

  

Question 65

A parallel-plate capacitor with the plate area 100cm2 and the separation between the plates 1.0cm is connected across a battery of emf 24 volts. Find the force of attraction between the plates.

Solution 65

  

C =   

Charge = Q = CV

=  

Q = 212.4  C

Force =   

=  

Force = 2.5 N

Question 66

Consider the situation shown in figure. The width of each plate is b. The capacitor plates are rigidly clamped in the laboratory and connected to a battery of emf. All surface are frictionless. Calculate the value of M for which the dielectric slab will stay in equilibrium.

 

  

Solution 66

Let the length of the part of slab inside the capacitor be x

 

 

Capacitance of air capacitor   =   

And dielectric capacitor   =   

Both are in parallel combination 

  

  

  

Energy stored =  

  

Force by capacitor on plate

  

  

  

Since it is in equilibrium

  

  

  

Question 67

Figure shows two parallel plate capacitors with fixed plates and connected to two batteries. The separation between the plates is the same for the two capacitors. The plates are rectangular in shape with width b and length l1 and l2. The left half of the dielectric slab has a dielectric constant K1 and the right half K2. Neglecting any friction, find the ratio of the emf of the left battery to that of the right battery for which the dielectric slab may remain in equilibrium.

 

 

Solution 67

 

For left capacitor ,   are in parallel combination

  

  

  

  

  

  

  

Similarly by right side capacitor

  

Now , dielectric is in equilibrium

  

  

Question 68

Consider the situation shown in figure. The plates of the capacitor have plate area A and are clamped in the laboratory. The dielectric slab is released from rest with a length a inside the capacitor. Neglecting any effect of friction or gravity, show that the slab will execute periodic motion and find its time period.

 

 

Solution 68

 

Let at any time dielectric is x length inside plates

Width of plate =  

Capacitance,

  

  

Both are in parallel combination

  

  

Force by capacitor

  

  

  

Negative shows that's force is in inside direction of capacitor.

Velocity of slabs increases till it comes completely inside the slab and then again decrease as when slab comes out force is inside the capacitor.

So, it performs periodic motion.

Time to come completely inside slab

U=0; s=(l-a); acc =  

  

  

  

For oscillation,

Time periods=4t