Electric Charges And Fields Class 12th Physics Part I CBSE Solution

Class 12th Physics Part I CBSE Solution
Exercises
  1. What is the force between two small charged spheres having charges of 2 × 10-7 C and 3 ×…
  2. The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of…
  3. Check that the ratio is dimensionless. Look up a Table of Physical Constants and determine…
  4. Explain the meaning of the statement ‘electric charge of a body is quantised’.…
  5. Why can one ignore quantisation of electric charge when dealing with macroscopic i.e.…
  6. When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon…
  7. Four-point charges qA = 2 μC, qB = -5 μC, qC = 2 μC, and qD = -5 μC are located at the…
  8. An electrostatic field line is a continuous curve. That is, a field line cannot have…
  9. Explain why two field lines never cross each other at any point?
  10. Two point charges qA = 3 μC and qB = -3 μC are located 20 cm apart in vacuum. (a) What is…
  11. A system has two charges qA = 2.5 × 10-7C and qB = -2.5 × 10-7 C located at points A: (0,…
  12. An electric dipole with dipole moment 4 × 10-9 C m is aligned at 30° with the direction of…
  13. A polythene piece rubbed with wool is found to have a negative charge of 3 × 10-7 C. (a)…
  14. Two insulated charged copper spheres A and B have their centres separated by a distance of…
  15. What is the force of repulsion if each sphere is charged double the above amount, and the…
  16. Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the…
  17. Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give…
  18. Consider a uniform electric field E = 3 × 10^3 î N/C. (a) What is the flux of this field…
  19. What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side…
  20. Careful measurement of the electric field at the surface of a black box indicates that the…
  21. A point charge + 10 μC is a distance 5 cm directly above the centre of a square of side 10…
  22. A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What…
  23. A point charge causes an electric flux of -1.0 × 10^3 Nm^2 /C to pass through a spherical…
  24. A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm…
  25. A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of…
  26. An infinite line charge produces a field of 9 × 10^4 N/C at a distance of 2 cm. Calculate…
  27. Two large, thin metal plates are parallel and close to each other. On their inner faces,…
Additional Exercises
  1. An oil drop of 12 excess electrons is held stationary under a constant electric field of…
  2. Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field…
  3. In a certain region of space, electric field is along the z-direction throughout. The…
  4. A conductor A with a cavity as shown in Fig. 1.36(a) is given a charge Q. Show that…
  5. A hollow charged conductor has a tiny hole cut into its surface. Show that the electric…
  6. Obtain the formula for the electric field due to a long thin wire of uniform linear charge…
  7. It is now believed that protons and neutrons (which constitute nuclei of ordinary matter)…
  8. Consider an arbitrary electrostatic field configuration. A small test charge is placed…
  9. A particle of mass m and charge (-q) enters the region between the two charged plates…
  10. Suppose that the particle in Exercise in 1.33 is an electron projected with velocity vx=…

Exercises
Question 1.

What is the force between two small charged spheres having charges of 2 × 10–7 C and 3 × 10–7C placed 30 cm apart in air?


Answer:

The coulomb attraction formula used to find the force, F is given as,

 …(1)

Where, q1 and q2 are the charges.

r is the distance between the charges.

 is a constant and its value is 9x109 N m2 C-2

ϵ0 is the permittivity of free space. Its value is 8.85 x 10-12 (F/m)

Now, the diagram is:


Since, both the charges are positive, thus, the nature of force will be repulsive. F12 is the force on charge q2 caused by charge q1 andF21 is the force on charge 1 due to charge 2 .

Now, Given:

Charge on the first sphere, q1 = 2 × 10–7 C

Charge on the second sphere, q2 = 3 × 10–7 C

Distance between the spheres, r(in m) =30/100=0.3 m

Putting the values in equation (1), we get,

F = 6 × 10-3 N

Hence, the force between the given charged particles will be 6 × 10-3 N. Since the nature of the charges is the same i.e. they are both positive. Hence, the force will be repulsive.


Question 2.

The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge –0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres? (b) What is the force on the second sphere due to the first?


Answer:

(a) The electrostatic force, F = 0.2 N


Charge on this present sphere, q1 = 0.4 μC = 0.4 × 10-6 C


Charge on the other sphere, q2 = –0.8 μC = -0.8 × 10-6 C


The electrostatic force between the spheres can be given by the equation,


 …(1)


Where, q1 and q2 are the charges.


r is the distance between the charges.


 is a constant and its value is 9x109 N m2 C-2


ϵ0 is the permittivity of free space. Its value is 8.85 x 10-12 (F/m)


Therefore, from equation (1),




⇒ r2 = 144 x 10-4 m2


Taking square root both the sides,


⇒ r = 


r = 0.12 m


Hence, the distance between the two spheres is 0.12 m.


(b) Since the nature of the charges on the spheres is opposite, the force between them will be attractive in nature. The spheres will attract each other with the same amount of force.


Therefore, the force on the second sphere due to the first sphere will be 0.2 N.


Question 3.

Check that the ratio  is dimensionless. Look up a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?


Answer:

The given ratio is 


It is the ratio of electric force i.e.to gravitational force i.e.


 between a proton and an electron with the distance between them as constant.


This is because


Here, G = Gravitational constant. (Unit = Nm2kg-2 )


me and mp are the mass of electron and proton respectively. (Unit = Kg)


e = Electric charge (Unit = Coulomb)


K is a constant = 


 is the permittivity of space. Its units are Nm2 C-2.


Let us substitute the units into the ratio in order to deduce the dimensions,



All the units get cancelled, and we see that the given ratio is dimensionless i.e. M0 L0 T0


Let us now calculate the value of the given ratio,


G = 6.67 × 1011 Nm2 Kg-2


e = 1.6 × 1019 C


me = 9.1 × 10-31 kg


mp = 1.66 × 10-27 kg


On substituting these values into the ratio, we get




Note:



This implies that the electric force between a proton and an electron is 1039 times greater than the gravitational force.


Question 4.

Explain the meaning of the statement ‘electric charge of a body is quantised’.


Answer:

Electric charge of a body is quantised.


This statement means that charge on a body can take only integral values i.e. (1, 2, 3, 4, …. N) number of electrons can be transferred from one body to another. The charges cannot be transferred in fractions. It can only be an integral multiple of the charge of one electron.


Therefore, as a result, a body will possess a charge that is an integral multiple of the electric charge of an electron.



Question 5.

Why can one ignore quantisation of electric charge when dealing with macroscopic i.e. large-scale charges?


Answer:

When considered on a macroscopic level or a large scale, we tend to ignore the quantisation of electric charges. This is because on a large scale, the charges that we deal with are extremely huge when compared to the magnitude of an individual charge. Hence, we can say that the quantisation of electric charge is not useful on a macroscopic scale. It is therefore ignored and considered to be of continuous nature.



Question 6.

When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.


Answer:

i. When two bodies are rubbed against each other, this process results in the production of charges on the bodies involved.


ii. The charges on each body will be of equal magnitude but of opposite nature. This happens because charges are always created in pairs.


This phenomenon of charging bodies by rubbing them against each other is termed as charging by friction. However, the net charge of the system – of both the bodies – is zero.


iii. This is because equal amounts of charge of opposite nature cancel each other out.


iv. When a glass rod is rubbed with a silk cloth, then opposite charges appear on both the bodies. This is in accordance with the law of conservation of energy. (Net charge of isolated system remains constant) This phenomenon is also observed when other pairs of bodies are also rubbed with each other.



Question 7.

Four-point charges qA = 2 μC, qB = –5 μC, qC = 2 μC, and qD = –5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 μC placed at the centre of the square?


Answer:

Consider a square of side 10 cm. Four charges are placed in its corners. Let O be the center of the square.


The diagram is:



AB = BC = CD = DA = 10 cm


Diagonal of a square = √2 × Side of a square


Therefore,


AC = BD = 10√2 cm


AO = OC = DO = OB = (10√2)/2 cm = 5√2cm


i. A charge of magnitude 1μC is placed at O, the center of the square.
ii. The force experienced at center due to side charge at B is same in magnitude but in a direction opposite to the direction to the force at center due to charge at D.
iii. The force experienced at center due to side charge at A is same in magnitude but in a direction opposite to the direction to the force at center due to charge at C .



Force on the 1 u C in the center is = F+ F+ F+ Fd = F+ (- Fa) + F+ ( - Fd) = 0

Therefore, we can conclude that the resultant force by the four charges kept in the corners on the charge that is kept in the center will be zero.


Question 8.

An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?


Answer:

An electric field line always points in the direction that a positive test charge would accelerate in if placed upon that line. A charge will experience a continuous force when placed in an electrostatic field. Therefore, an electrostatic field line will be a continuous curve. The field lines do not have any sudden breaks as the test charge or any charge will move continuously and does not jump from any point to another.



Question 9.

Explain why two field lines never cross each other at any point?


Answer:

An electric field line represents the direction in which a positive test charge would accelerate in if placed on the line. If two field lines intersect, then this will imply that the electric field intensity points to two directions at the same point. This is not possible.


Hence, two electric field lines will never cross each other.



Question 10.

Two point charges qA = 3 μC and qB = –3 μC are located 20 cm apart in vacuum.

(a) What is the electric field at the midpoint O of the line AB joining the two charges?

(b) If a negative test charge of magnitude 1.5 × 10–9 C is placed at this point, what is the force experienced by the test charge?


Answer:

(a) The diagram is shown as below:



AB = 20 cm


∵ O is the midpoint of the line AB.


∴ AO = OB = 10 cm = 0.1m


Let the Net electric field at the point O = E


The electric field at a point caused by charge q, is given as,


 …(1)


Where, q is the charge,


r is the distance between the charge and the point at which the field is being calculated


 is a constant and its value is 9x109 N m2 C-2


ϵ0 is the permittivity of free space. Its value is 8.85 x 10-12 (F/m)


∴ The electric field at the point O caused by qA = 3 μC




The direction of EA will be along OB.


Also, the electric field at O caused by qB = –3 μC




The direction of EB will be along OB.


Note: Both the field are acting in the same direction along OB, this is because the charges are of opposite nature, ∴ the force will be attractive.


Hence, we can sum them up to find the resultant electric field.


Enet = EA + EB


⇒ Enet = 2EA



∴ Enet = 5.4 × 102 N/C(along OB)


(b) The diagram is:



q = 1.5 × 10–9 C


The force that is experienced by q when placed at O, F.


FO = q × ENet


Where E is the Electric field at the point O.


∴ F = 1.5 × 10–9 C x 5.4 × 102 N/C


⇒ F = 8.1 x 10-7 N


The negative test charge will be repelled by the force placed at B and attracted by the force placed at A. Therefore, this force will be in the direction of OA.


Question 11.

A system has two charges qA = 2.5 × 10–7C and qB = –2.5 × 10–7 C located at points A: (0, 0, –15 cm) and B: (0,0, + 15 cm), respectively. What are the total charge and electric dipole moment of the system?


Answer: 


The charge at point A, qA = 2.5 × 10–7C


The charge at point B, qB = –2.5 × 10–7 C


∴ the total charge of the system, Qnet = qA + qB


Qnet = 2.5 × 10–7C + (–2.5 × 10–7 C)


Qnet = 0


The distance between these two charges = Distance between point A and B, d = 15 + 15


d = 30 cm = 0.3 m


The electric dipole moment can be defined as the measure of the separation of positive as well as negative charges within a specific system. It tells us about the given system’s overall polarity.


The electric dipole moment of the system, p = qA x d = qb x d


p = 2.5 × 10–7C x 0.3


∴ p = 7.5 x 10-8 C-m along the positive z-axis


Hence, the electric dipole moment of this system is 7.5 x 10-8 C-m along the positive z axis.


Question 12.

An electric dipole with dipole moment 4 × 10–9 C m is aligned at 30° with the direction of a uniform electric field of magnitude 5 × 104 NC–1. Calculate the magnitude of the torque acting on the dipole.


Answer:

The electric dipole moment can be defined as the measure of the separation of positive as well as negative charges within a specific system. It tells us about the given system’s overall polarity.


Electric dipole moment, p = 4 × 10–9 C-m


Angle between p and uniform electric field,  = 30


Electric field, E = 5 × 104 NC–1


The torque acting on a dipole is given by:


τ = p × E


τ = p×E×sinθ(θ=30)


= 4 × 10–9 C m x 5 × 104 NC–1 x (0.5)

= 10-4 N m

The magnitude of torque acting on the given dipole is 10-4 N m.


Question 13.

A polythene piece rubbed with wool is found to have a negative charge of 3 × 10–7 C.

(a) Estimate the number of electrons transferred (from which to which?)

(b) Is there a transfer of mass from wool to polythene?


Answer:

Given:


Charge on polythene piece = -3 × 107 C


(a) Number of electrons transferred from wool to plastic.


Explanation: We know that all materials are electrically neutral. As per the question, after rubbing the polythene piece is found to be negatively charged which suggests that negatively charged particles i.e. electrons were transferred from wool to plastic.


Let number of electrons transferred from wool to plastic be ‘n’


From quantization of charges we know that


q = n × e …(1)


Where, q is the total charge


e = charge of individual particle


n = number of charges (either positive or negative)


e = -1.6 × 10-19 C


From equation (1) we get,


q = n × e


⇒ n = q/e


⇒ 


⇒ n = 1.87 × 1012


The number of electrons transferred is 1.8 × 1012


(b) Yes, mass is transferred because each electron has a mass of 9.1 × 10-31 Kg.


Mass transferred = n × me …(1)


Where, n = number of electrons


Me = mass of electron (9.1 × 10-31Kg)


Now, putting the values of Me and n in equation (1).


⇒ M = 1.87 × 1012 × 9.1 × 10-31Kg


⇒ M = 1.706 × 10-18 Kg


Hence, the mass transferred from wool to plastic is 1.7 × 10-18 Kg



Question 14.

Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 × 10–7 C? The radii of A and B are negligible compared to the distance of separation.


Answer:

Given:


Distance between centres of spheres, r = 50 cm = .5m


Charge on each sphere, q = 6.5 × 10-7 C



Mutual force of electrostatic repulsion


…(1)


Where, F= mutual force of attraction


q1 = charge on sphere 1


q2 = charge on sphere 2


r = distance between centres



Where, ε0 is the permittivity of the free space.


Now, putting the values of q1, q2 and r in equation (1).



⇒ F = 1.52 × 10-2 N


Hence the mutual force between two spheres is 1.52 × 10-2N. Since the sign is positive so the force is repulsive in nature.



Question 15.

What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?


Answer:

Force if charges on each sphere is doubled and the distance is halved.


Given:


Q1 = 2 × q


Q2 = 2 × q


R = 0.5 × r


Using Coulomb’s law, we get,


…(1)



Where, ε0 is the permittivity of the free space.


⇒ 


⇒ 


⇒ F = 0.243 N


The mutual force is 0.243N and it is repulsive in nature.


The new force is found to be 16 times the force in case I.



Question 16.

Suppose the spheres A and B in Exercise 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?


Answer:

Note: If a conducting uncharged material is brought in contact with a charged surface then the charges are shared uniformly between the two bodies.


Given: Charge on sphere 1, q1 = 6.5 × 10-7 C


Charge on sphere 2, q2 = 6.5 × 10-7 C


Charge on sphere 3, q3 = 0



Step 1: The uncharged sphere is brought in contact with sphere 1. Since sphere 1 has charge ‘q’, it gets distributed among sphere 1 and sphere 3.


Now, charge on sphere 1 = q/2


Charge on sphere 2 = q/2


At this point the sphere 3 which was initially uncharged has a charge “q/2”.


Step 2: Now sphere 3 is brought in contact with sphere 2 due to which 1/4 × q will flow from sphere 2 to sphere 3. Now sphere 2 and sphere 3 have “3/4 × q” charge.


Now, q1 = 1/2 × 6.5 × 10-7 C


⇒ 3.25 × 10-7 C


q2 = 3/4 × 6.5 × 10-7 C


⇒ 4.87 × 10-7 C


q3 = 3/4 × 6.5 × 10-7 C


⇒ 4.87 × 10-7 C


We know that,


…(1)


Where, F= mutual force of attraction


q1 = charge on sphere 1


q2 = charge on sphere 2


r = distance between centres



Where, ε0 is the permittivity of the free space.


Plugging the values of q1, q2 and r in equation (1), we get


⇒ 


⇒ F = 5.703 × 10-3 N


The repulsive force between sphere 1 and sphere 2 is 5.703 × 10-3 N.



Question 17.

Figure 1.33 shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?



Answer:

1. Particles 1 and 2 are deflected towards the positively charged plate, hence it can be concluded that particles 1 and 2 are negatively charged.


2. Particle 3 is deflected towards the negatively charged plate hence it is positively charged.


The deflection of a particle is proportional to the charge to mass ratio (q: m) of the particle. We observe that particle 3 has maximum deflection, so it has the highest charge to mass ratio.


Explanation: Force on a particle in electric field


F = q × E …(1)


Where, q = charge on particle


E = magnitude of electric field


m × a = q × E …(2)


Where m = mass of particle


a = acceleration of particle


From equation (2)


⇒ 


The electric field in the given case is constant, hence we can write


⇒  …(3)


From the equation (3) we can conclude that acceleration of a charge particle with charge q and mass m in uniform electric field is proportional to the charge to mass ratio of the particle.



Question 18.

Consider a uniform electric field E = 3 × 103î N/C. (a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane? (b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?


Answer:

Given:


Electric field E = 3 × 103 N/C


Side of square, s = 10 cm



a) Flux of field through square whose plane is parallel to yz plane.


We understand that the normal to the plane is parallel to the direction of field.


So, θ = 0°


Φ = E . A


Φ = E × A × cos(θ) …(1)


Where, E = Electric field


A = Area through which we have to calculate flux


θ = Angle between normal to surface and the Electric field


A = s2


A = .01 m2


Plugging values, of E, A and θ in equation (1)


Φ = 3 × 103 NC-1 × 0.01m2 × cos0°


Φ = 30 Nm2C-1


b) If normal to its (square’s) plane makes 60° with the X axis.


θ = 60°


Φ = E × A × cos(θ)


Φ = 3 × 103 NC-1 × 0.01m2 × cos60°


Φ = 15 Nm2C-1



Question 19.

What is the net flux of the uniform electric field of Exercise 1.15 through a cube of side 20 cm oriented so that its faces are parallel to the coordinate planes?


Answer:

Flux, Φ = q/ε0


Where, q = net charged enclosed


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


Since no charge is enclosed by the cube, i.e. q = 0, the net flux through the Cube is zero.



Note: The number of field lines entering the cube is equal to the number of field lines leaving the cube. Hence the net flux becomes zero.



Question 20.

Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 Nm2/C. (a) What is the net charge inside the box? (b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or Why not?


Answer:

Given:

(a) Φ = 8.0 × 103 Nm2C-1


Let net charge inside the box = q


We know that,


Flux, Φ = q/ε0 ..(1)


Where, q = net charged enclosed


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


Plugging values of Φ and ε0 in equation (1) we get,


q = Φ × ε0


⇒ q = 8.0 × 103 Nm2C-1 × 8.85 × 10-12N-1C2m-2


⇒ q = 7.08 × 10-8 C


Hence, the net charge inside the box is 0.07 μC.


b) No, we cannot conclude that the body doesn’t have any charge. The flux is due to the Net charge of the body. There may still be equal amount of positive and negative charges. So, it is not necessary that if flux is zero then there will be no charges.



Question 21.

A point charge + 10 μC is a distance 5 cm directly above the centre of a square of side 10 cm, as shown in Fig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one face of a cube with edge 10 cm.)



Answer:

Given:


q = + 10 


s = 10 cm


Assume the charge to be enclosed by a cube, where the square is one of its sides.


Now, let us find the total flux through the imaginary cube.


We know that,


Flux, Φ = q/ε …(1)


Where, q = net charged enclosed


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


Now plugging the values of q and ε0 in equation (1)



⇒ Φ = 11.28 × 105 Nm-2C-1


We understand that flux through all the faces of cube will be equal;


Let flux through the square = Φa


Hence,


Φa = Φ/6


Explanation: The net flux will be distributed equally among all 6 faces of the cube. Hence, the square will have one sixth of the total flux.


Φa = 1.88 Nm-2C-1


The flux through the square is.


Question 22.

A point charge of 2.0 μC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?


Answer:

Given:


Total charge inside the cube, q = 2.0 μC


Edge length of Cube, a = 9.0 cm



We know that,


Flux, Φ = q/ε0


Where, q = net charged enclosed


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2



⇒ Φ = 2.26 × 105 Nm2C-1


The net electric flux through the cubic Gaussian surface is Φ = 2.26 × 105 Nm2C-1


NOTE: The flux through a Gaussian surface depends on the net charge enclosed by it. Geometry has no effect on the total flux.



Question 23.

A point charge causes an electric flux of –1.0 × 103 Nm2/C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?


Answer:

Given:


Φ = -1.0 × 103 Nm2C-1


r1 = 10.0 cm.



a) Flux if the radius of the Gaussian surface is doubled.


If the radius is doubled then the flux would remain same i.e. -1.0 × 103 Nm2C-1.


The geometry of the Gaussian surface doesn’t affect the total flux through it. The net charge enclosed by Gaussian surface determines the net flux.


b) Let value of point charge enclosed by Gaussian surface = q


We know that,


Flux, Φ = q/ε0 …(1)


Where, q = net charged enclosed


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


Now plugging, the values Φ and ε0 in equation (1).


q = Φ × ε0


⇒ q = -1.0 × 103Nm2C-1 × 8.85 × 10-12N-1m-2C2


⇒ q = -8.85 × 10-9 C


The charge enclosed by the surface is -8.85 × 10-9 C.



Question 24.

A conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the centre of the sphere is 1.5 × 103 N/C and points radially inward, what is the net charge on the sphere?


Answer:

Given:


Radius of charged sphere, r = 10 cm


Electric field, 20 cm away from centre of sphere, E = 1.5 × 103 NC-1


We know that, electric field intensity at a point P, located at a distance R, due to net charge q is given by,


 …(1)


Now plugging the values of q and R in equation (1)


q = 4 × π × ε0 × R2 × E


⇒ q = 4 × 3.14 × 8.85 × 10-12N-1m-2C-2 × (0.2m)2 × 1.5 × 103C


⇒ q = 6.67 × 10-9C


The net charge on the sphere is -6.67 × 10-9C. Since the electric field points radially inwards, we can infer that charges on sphere are negative.



Question 25.

A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.0 μC/m2. (a) Find the charge on the sphere. (b) What is the total electric flux leaving the surface of the sphere?


Answer:

Given: radius of sphere, r = 1.2 m


Surface charge density, σ = 80.0 μC/m2



a) Let charge on sphere = q


We understand that,


Total charge, Q = Surface charge density × surface area.


Surface area of sphere, S = 4πr2


S = 18.08 m2


Q = 80 × 10-6Cm-2 × 18.08m-2


⇒ Q = 1.447 × 10-3C


b) Let total electric flux leaving the surface of sphere = 


We know that,


Flux, Φ = q/ε0 …(1)


Where, Q = net charged.


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


By, plugging the values of q and ε0 in equation (1), we get,


⇒ 


⇒ Φ = 1.63 × 108 Nm2C-1


The total flux through the sphere is 1.63 × 108 Nm2C-1 .



Question 26.

An infinite line charge produces a field of 9 × 104 N/C at a distance of 2 cm. Calculate the linear charge density.


Answer:

Given:


E = 9 × 104 NC-1


d = 2 cm


Let the linear charge density = λ Coulomb/metre



We know that, Electric field produced by a line charge with a linear charge density σ, at a distance d is given by,


 …(1)


Where, λ = linear charge density.


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


d = distance


From equation (1) we have,


⇒ λ = E × 2 × π × ε0 × d


⇒ λ = 9 × 104 NC-1 × 2 × π × 8.85 × 10-12N-1 m-2C2 × 0.02m


⇒ λ = 10 ×10-8 Cm-1


The linear charge density is 10-7 Cm-1.


Question 27.

Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 × 10–22 C/m2. What is E: (a) in the outer region of the first plate, (b) in the outer region of the second plate, and (c) between the plates?


Answer:

Given:


Surface charge density on plate A, σA = -17.0 × 10-22 Cm-2


Surface charge density on plate B, σB = 17.0 × 10-22 Cm-2


The arrangement of plates are as shonw:



Let electric field in region 1 = E1


Region 2 = E2


Region 3 = E3


The electric field in region I and region III is zero because no charge is present in these regions.


E1 = 0


E3 = 0


We know that,


E2 = σ/ε0 ...(1)


Where, σ Surface charge density


ε0 = permittivity of free space


ε0 = 8.85 × 10-12N-1 m-2C2


Now, plugging the Values in equation (1).


⇒ 


⇒ E2 = 1.92 × 10-10 NC-1


The electric field in the region enclosed by the plates is found to be 1.92 × 10-10 NC-1.


The electric field in region II is 1.92 × 10-10 NC-1.



Additional Exercises
Question 1.

An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 × 104 NC–1 in Millikan’s oil drop experiment. The density of the oil is 1.26 gcm-3. Estimate the radius of the drop.

(g = 9.81 m s–2; e = 1.60 × 10–19 C).


Answer:

If the oil drop is stationary, the net force on it must be Zero or the resultant of all the forces on oil drop is zero.


There are two forces acting on the oil drop, its weight or force due to Earth’s gravity which is pulling it vertically downwards, and Electrostatic force which is acting in vertically upward direction.


Both forces should be equal in magnitude and opposite direction so that they cancel each other.


The arrangement is shown in the figure



Note: Electric Field is in vertically downward direction, because force on an negatively charged body in opposite direction of the field, so force on the drop is in vertically upward direction, which balances weight of the drop acting in vertically downward direction.


There are nine excess electrons which make the drop negatively charged because the electron is negatively charged, the net magnitude charge on a body is given by


q = n × E


n is the excess of electrons or protons on the body,


the charge of an electron is denoted by e


e = 1.60 × 10–19 C


here since there are twelve excess electrons so


n = 12


i.e. q = 12 × 1.60 × 10–19 C


= 1.92 × 10-18 C


Now the magnitude electrostatic force on a charged particle held in an electric field is given by


F = q × E


where, F force is acting on a particle having charge q held in an electric field of magnitude E


here, charge on the oil drop is


q = 1.92 × 10-18 C


magnitude of Electric field is


E = 2.55 × 104NC–1


So Electrostatic force on the oil drop is


F = 1.92 × 10-18 C × 2.55 × 104NC–1


= 4.896 × 10-14 N


This force is acting in vertically upward direction, so Weight should have same magnitude of Force and is acting in vertically downward direction.


Let us assume oil drop to be Spherical in shape, so the volume of drop will be


 (Volume of Sphere)


Where r is the radius of the Spherical drop


Now we know mass of an object whose volume and density are known is given by


m = V × d


where, m is the mass


V is the volume and d is the density


Here density of the drop is


d = 1.26 gcm-3


= 1.26 × 103 Kgm-3


so mass of the drop is


m =  × 1.26 × 103


Now the downward pull on a body due to earth’s gravitational force is the weight of body given by


W = m × g


Where W is the weight of a body having mass m


The acceleration due to gravity is denoted by g


g = 9.81 ms-2


so the weight or the downward gravitational force on oil drop is


W =  × 1.26 × 103 Kgm-3× 9.81ms-2


Both the forces should be equal in magnitude so equating them


i.e. putting F = W


4.896 × 10-14 N = (4/3 πr3) × 1.26 × 103 Kgm-3 × 9.81ms-2


we get ,


(1N = 1kgms-2)



So, r 



⇒r = 0.98110-6 m


= 9.81 × 10-7 m


= 981 × 10-4 mm


i.e. radius of the oil drop is 98110-4 mm



Question 2.

Which among the curves shown in Fig. 1.35 cannot possibly represent electrostatic field lines?





Answer:

Electrostatic field lines have some properties and characteristics, which should be taken into account while drawing Electrostatic field lines, the Characteristic of electrostatic field lines are :-


1. They start from a positively charged body and end at a negatively charged body.


2. Tangent to the electrostatic field line at any point gives the direction of the electric intensity at that point.


3. Electrostatic field lines cannot intersect each other.


4. The electrostatic field lines are always normal to the surface of a conductor. There is no component of the electric filed intensity parallel to the surface of the conductor.


5. Electrostatic field lines do not form any closed loops.


These properties and characteristics must be taken into account while drawing electrostatic field lines, if any of these are violated then the following representation will be incorrect.