Chapter 1 Electric Charges And Fields Solutions
Question - 1 : - What is the force between twosmall charged spheres having charges of 2 × 10−7 C and 3 × 10−7 Cplaced 30 cm apart in air?
Answer - 1 : -
Repulsive force of magnitude 6 ×10−3 N
Charge on the first sphere, q1 =2 × 10−7 C
Charge on the secondsphere, q2 = 3 × 10−7 C
Distance between thespheres, r = 30 cm = 0.3 m
Electrostatic force between thespheres is given by the relation,
Where, ∈0 =Permittivity of free space
Hence, force between the twosmall charged spheres is 6 × 10−3 N. The charges are of samenature. Hence, force between them will be repulsive.
Question - 2 : - The electrostatic force on asmall sphere of charge 0.4 μC due to another small sphere of charge − 0.8 μC inair is 0.2 N. (a) What is the distance between the two spheres? (b) What is theforce on the second sphere due to the first?
Answer - 2 : -
(a) Electrostaticforce on the first sphere, F = 0.2 N
Charge on this sphere, q1 =0.4 μC = 0.4 × 10−6 C
Charge on the secondsphere, q2 = − 0.8 μC = − 0.8 × 10−6 C
Electrostatic force between thespheres is given by the relation,
Where, ∈0 =Permittivity of free space
The distance between the twospheres is 0.12 m.
(b) Both thespheres attract each other with the same force. Therefore, the force on thesecond sphere due to the first is 0.2 N.
Question - 3 : - Check that the ratio ke2/Gmemp is dimensionless. Look up aTable of Physical Constants and determine the value of this ratio. What doesthe ratio signify?
Answer - 3 : - The given ratio is 
Where,
G = Gravitational constant
Its unit is N m2 kg−2.
me and mp =Masses of electron and proton.
Their unit is kg.
e =Electric charge.
Its unit is C.
∈0 = Permittivity of free space
Its unit is N m2 C−2.
Hence, the given ratio isdimensionless.
e = 1.6 ×10−19 C
G = 6.67 × 10−11 Nm2 kg-2
me= 9.1 × 10−31 kg
mp =1.66 × 10−27 kg
Hence, the numerical value of thegiven ratio is
This is the ratio of electricforce to the gravitational force between a proton and an electron, keepingdistance between them constant.
(a) Explainthe meaning of the statement ‘electric charge of a body is quantised’.
(b) Whycan one ignore quantisation of electric charge when dealing with macroscopici.e., large scale charges?
Answer - 4 : -
(a) Electriccharge of a body is quantized. This means that only integral (1, 2, …., n)number of electrons can be transferred from one body to the other. Charges arenot transferred in fraction. Hence, a body possesses total charge only inintegral multiples of electric charge.
(b) In macroscopic or largescale charges, the charges used are huge as compared to the magnitude ofelectric charge. Hence, quantization of electric charge is of no use onmacroscopic scale. Therefore, it is ignored and it is considered that electriccharge is continuous
Question - 5 : - When a glass rod is rubbed with asilk cloth, charges appear on both. A similar phenomenon is observed with manyother pairs of bodies. Explain how this observation is consistent with the lawof conservation of charge.
Answer - 5 : -
Rubbing produces charges of equalmagnitude but of opposite nature on the two bodies because charges are createdin pairs. This phenomenon of charging is called charging by friction. The netcharge on the system of two rubbed bodies is zero. This is because equal amountof opposite charges annihilate each other. When a glass rod is rubbed with asilk cloth, opposite natured charges appear on both the bodies. This phenomenonis in consistence with the law of conservation of energy. A similar phenomenonis observed with many other pairs of bodies.
Question - 6 : - Four point charges qA =2 μC, qB = −5 μC, qC =2 μC, and qD = −5 μC are located at the corners ofa square ABCD of side 10 cm. What is the force on a charge of 1 μC placed atthe centre of the square?
Answer - 6 : -
The given figure shows a squareof side 10 cm with four charges placed at its corners. O is the centre of thesquare.
Where,
(Sides) AB = BC = CD = AD = 10 cm
(Diagonals) AC = BD = 
cm
AO = OC = DO = OB = 
cm
A charge of amount 1μC is placedat point O.
Force of repulsion betweencharges placed at corner A and centre O is equal in magnitude but opposite indirection relative to the force of repulsion between the charges placed atcorner C and centre O. Hence, they will cancel each other. Similarly, force ofattraction between charges placed at corner B and centre O is equal inmagnitude but opposite in direction relative to the force of attraction betweenthe charges placed at corner D and centre O. Hence, they will also cancel eachother. Therefore, net force caused by the four charges placed at the corner ofthe square on 1 μC charge at centre O is zero.
Question - 7 : - (a) Anelectrostatic field line is a continuous curve. That is, a field line cannothave sudden breaks. Why not?
(b) Explainwhy two field lines never cross each other at any point?
Answer - 7 : -
(a) Anelectrostatic field line is a continuous curve because a charge experiences acontinuous force when traced in an electrostatic field. The field line cannothave sudden breaks because the charge moves continuously and does not jump fromone point to the other.
(b) Iftwo field lines cross each other at a point, then electric field intensity willshow two directions at that point. This is not possible. Hence, two field linesnever cross each other.
Question - 8 : - Two point charges qA =3 μC and qB = −3 μC are located 20 cm apart invacuum.
(a) Whatis the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative testcharge of magnitude 1.5 × 10−9 C is placed at this point, whatis the force experienced by the test charge?
Answer - 8 : -
(a) Thesituation is represented in the given figure. O is the mid-point of line AB.
Distance between the two charges,AB = 20 cm
∴AO = OB = 10 cm
Net electric field at point O = E
Electricfield at point O caused by +3μC charge,
E1 =
along OB
Where,
Permittivity of free space
Magnitude of electric field atpoint O caused by −3μC charge,
E2 = 
= 
along OB
= 5.4 × 106 N/Calong OB
Therefore, the electric field atmid-point O is 5.4 × 106 N C−1 along OB.
(b) A testcharge of amount 1.5 × 10−9 C is placed at mid-point O.
q =1.5 × 10−9 C
Force experienced by the testcharge = F
∴F = qE
= 1.5 × 10−9 ×5.4 × 106
= 8.1 × 10−3 N
The force is directed along lineOA. This is because the negative test charge is repelled by the charge placedat point B but attracted towards point A.
Therefore, the force experiencedby the test charge is 8.1 × 10−3 N along OA.
Question - 9 : - A system has two charges qA =2.5 × 10−7 C and qB = −2.5 × 10−7 Clocated at points A: (0, 0, − 15 cm) and B: (0, 0, + 15 cm), respectively. Whatare the total charge and electric dipole moment of the system?
Answer - 9 : -
Both the charges can be locatedin a coordinate frame of reference as shown in the given figure.
At A, amount of charge, qA =2.5 × 10−7C
At B, amount of charge, qB =−2.5 × 10−7 C
Total charge of the system,
q = qA + qB
= 2.5 × 10−7 C −2.5 × 10−7 C
= 0
Distance between two charges atpoints A and B,
d =15 + 15 = 30 cm = 0.3 m
Electric dipole moment of thesystem is given by,
p = qA × d= qB × d
= 2.5 × 10−7 ×0.3
= 7.5 × 10−8 C malong positive z-axis
Therefore, the electric dipolemoment of the system is 7.5 × 10−8 C m along positive z−axis.
Question - 10 : - An electric dipole with dipolemoment 4 × 10−9 C m is aligned at 30° with the direction of auniform electric field of magnitude 5 × 104 N C−1.Calculate the magnitude of the torque acting on the dipole.
Answer - 10 : -
Electric dipole moment, p =4 × 10−9 C m
Angle made by p witha uniform electric field, θ = 30°
Electric field, E =5 × 104 N C−1
Torque acting on the dipole isgiven by the relation,
τ= pE sinθTherefore, the magnitude of thetorque acting on the dipole is 10−4 N m.