Chapter 1 Electric Charges And Fields Solutions
Question - 11 : - A polythene piece rubbed withwool is found to have a negative charge of 3 × 10−7 C.
(a) Estimatethe number of electrons transferred (from which to which?)
(b) Isthere a transfer of mass from wool to polythene?
Answer - 11 : -
(a) Whenpolythene is rubbed against wool, a number of electrons get transferred fromwool to polythene. Hence, wool becomes positively charged and polythene becomesnegatively charged.
Amount of charge on the polythenepiece, q = −3 × 10−7 C
Amount of charge on anelectron, e = −1.6 × 10−19 C
Number of electrons transferredfrom wool to polythene = n
n can becalculated using the relation,
q = ne= 1.87 × 1012
Therefore, the number ofelectrons transferred from wool to polythene is 1.87 × 1012.
(b) Yes.
There is a transfer of masstaking place. This is because an electron has mass,
me =9.1 × 10−3 kg
Total mass transferred topolythene from wool,
m = me ×n
= 9.1 × 10−31 ×1.85 × 1012
= 1.706 × 10−18 kg
Hence, a negligible amount ofmass is transferred from wool to polythene.
(a) Twoinsulated charged copper spheres A and B have their centers separated by adistance of 50 cm. What is the mutual force of electrostatic repulsion if thecharge on each is 6.5 × 10−7 C? The radii of A and B arenegligible compared to the distance of separation.
(b) What is the force ofrepulsion if each sphere is charged double the above amount, and the distancebetween them is halved?
Answer - 12 : -
(a) Charge onsphere A, qA = Charge on sphere B, qB =6.5 × 10−7 C
Distance between thespheres, r = 50 cm = 0.5 m
Forceof repulsion between the two spheres,Where,
∈0 = Free space permittivity
= 9 × 109 N m2 C−2
∴
= 1.52 × 10−2 N
Therefore, the force between thetwo spheres is 1.52 × 10−2 N.
(b) Afterdoubling the charge, charge on sphere A, qA =Charge on sphere B, qB = 2 × 6.5 × 10−7 C= 1.3 × 10−6 C
The distance between the spheresis halved.
∴
Force of repulsion between thetwo spheres,
= 16 × 1.52 × 10−2
= 0.243 N
Therefore, the force between thetwo spheres is 0.243 N.
Suppose the spheres A and B inExercise 1.12 have identical sizes. A third sphere of the same size butuncharged is brought in contact with the first, then brought in contact withthe second, and finally removed from both. What is the new force of repulsionbetween A and B?
Answer - 13 : -
Distance between the spheres, Aand B, r = 0.5 m
Initially, the charge on eachsphere, q = 6.5 × 10−7 C
Whensphere A is touched with an uncharged sphere C, 
amount of charge from A will transfer to sphereC. Hence, charge on each of the spheres, A and C, isWhen sphere C with charge is brought in contact withsphere B with charge q, total charges on the system will divideinto two equal halves given as, Each sphere will share each half. Hence, charge oneach of the spheres, C and B, is
.
Force of repulsion between sphere A havingcharge and sphere B having charge .
Therefore, the force of attractionbetween the two spheres is 5.703 × 10−3 N.
Question - 14 : - Figure 1.33 shows tracks of threecharged particles in a uniform electrostatic field. Give the signs of the threecharges. Which particle has the highest charge to mass ratio?
Answer - 14 : -
Opposite charges attract eachother and same charges repel each other. It can be observed that particles 1and 2 both move towards the positively charged plate and repel away from thenegatively charged plate. Hence, these two particles are negatively charged. Itcan also be observed that particle 3 moves towards the negatively charged plateand repels away from the positively charged plate. Hence, particle 3 ispositively charged.
Thecharge to mass ratio (emf) is directly proportional to the displacement oramount of deflection for a given velocity. Since the deflection of particle 3is the maximum, it has the highest charge to mass ratio.
Question - 15 : - Consider a uniform electricfield E = 3 × 103 îN/C.(a) What is the flux of this field through a square of 10 cm on a side whoseplane is parallel to the yz plane? (b) What is the fluxthrough the same square if the normal to its plane makes a 60° angle withthe x-axis?
Answer - 15 : - (a) Electricfield intensity, 
= 3 × 103 îN/C
Magnitudeof electric field intensity, 
= 3 × 103 N/C
Side of the square, s =10 cm = 0.1 m
Area of the square, A =s2 = 0.01 m2
The plane of the square isparallel to the y-z plane. Hence, angle between the unitvector normal to the plane and electric field, θ = 0°
Flux (Φ) through the planeis given by the relation,
Φ= 
= 3 × 103 × 0.01× cos0°
= 30 N m2/C
(b) Planemakes an angle of 60° with the x-axis. Hence, θ =60°
Flux, Φ == 3 × 103 × 0.01× cos60°
= 15 N m2/C
What is the net flux ofthe uniform electric field of Exercise 1.15 through a cube of side 20 cmoriented so that its faces are parallel to the coordinate planes?
Answer - 16 : -
All the faces of a cubeare parallel to the coordinate axes. Therefore, the number of field linesentering the cube is equal to the number of field lines piercing out of thecube. As a result, net flux through the cube is zero.
Question - 17 : - Careful measurement of theelectric field at the surface of a black box indicates that the net outwardflux through the surface of the box is 8.0 × 103 Nm2/C. (a) What is the net chargeinside the box? (b) If the net outward flux through the surface of the box werezero, could you conclude that there were no charges inside the box? Why or Whynot?
Answer - 17 : -
(a) Net outward flux throughthe surface of the box, Φ = 8.0× 103 N m2/C
For a body containing netcharge q, flux is given by therelation,
∈0 = Permittivity offree space
= 8.854 × 10−12 N−1C2 m−2
q = ∈0Φ
= 8.854 × 10−12 × 8.0 × 103
= 7.08 × 10−8
= 0.07 μC
Therefore, the net chargeinside the box is 0.07 μC.
(b) No
Net flux piercing outthrough a body depends on the net charge contained in the body. If net flux iszero, then it can be inferred that net charge inside the body is zero. The bodymay have equal amount of positive and negative charges.
Question - 18 : - A point charge +10 μC is adistance 5 cm directly above the centre of a square of side 10 cm, as shown inFig. 1.34. What is the magnitude of the electric flux through the square? (Hint: Think of the square as one faceof a cube with edge 10 cm.)
Answer - 18 : -
The square can beconsidered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’stheorem for a cube, total electric flux is through all its six faces.
Hence, electric flux through one face of the cubei.e., through the square,
Where,
∈0 = Permittivity offree space
= 8.854 × 10−12 N−1C2 m−2
q = 10 μC = 10 × 10−6 C
∴
= 1.88 × 105 N m2 C−1
Therefore, electric fluxthrough the square is 1.88 × 105 Nm2 C−1.
Question - 19 : - A point charge of 2.0 μCis at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the netelectric flux through the surface?
Answer - 19 : -
Net electric flux (ΦNet) through the cubic surface is given by,
Where,
∈0 = Permittivity offree space
= 8.854 × 10−12 N−1C2 m−2
q = Net charge contained inside the cube = 2.0 μC = 2 × 10−6 C
∴
= 2.26 × 105 N m2 C−1
The net electric fluxthrough the surface is 2.26 ×105 Nm2C−1.
Question - 20 : - A point charge causes anelectric flux of −1.0 × 103 Nm2/C to pass through a spherical Gaussiansurface of 10.0 cm radius centered on the charge. (a) If the radius of theGaussian surface were doubled, how much flux would pass through the surface?(b) What is the value of the point charge?
Answer - 20 : -
(a) Electric flux, Φ = −1.0 × 103 Nm2/C
Radius of the Gaussiansurface,
r = 10.0 cm
Electric flux piercing outthrough a surface depends on the net charge enclosed inside a body. It does notdepend on the size of the body. If the radius of the Gaussian surface isdoubled, then the flux passing through the surface remains the same i.e., −103 N m2/C.
(b) Electric flux is given bythe relation,
Where,
q = Net charge enclosed by the spherical surface
∈0 = Permittivity offree space = 8.854 × 10−12 N−1C2 m−2
∴
= −1.0 × 103 × 8.854 × 10−12
= −8.854 × 10−9 C
= −8.854 nC
Therefore, the value ofthe point charge is −8.854 nC.