Chapter 1 Electric Charges And Fields Solutions
Question - 31 : - It is now believed thatprotons and neutrons (which constitute nuclei of ordinary matter) arethemselves built out of more elementary units called quarks. A proton and aneutron consist of three quarks each. Two types of quarks, the so called ‘up’quark (denoted by u) of charge (+2/3) e, and the ‘down’ quark(denoted by d) of charge (−1/3) e, together with electrons build upordinary matter. (Quarks of other types have also been found which give rise todifferent unusual varieties of matter.) Suggest a possible quark composition ofa proton and neutron.
Answer - 31 : - A proton has three quarks. Let there be n upquarks in a proton, each having a charge of 
Charge due to n up quarks
Number of down quarks in aproton = 3 − n
Eachdown quark has a charge of 
Charge due to (3 − n) down quarks 
Total charge on a proton =+ e
Number of up quarks in aproton, n = 2
Number of down quarks in aproton = 3 − n = 3 − 2 = 1
Therefore, a proton can berepresented as ‘uud’.
Aneutron also has three quarks. Let there be n up quarks in aneutron, each having a charge of 
Charge on a neutron due to n upquarks 
Number of down quarks is 3 − n,each havinga charge of
Charge on a neutron due to 
down quarks = 
Total charge on a neutron= 0
Number of up quarks in aneutron, n = 1
Number of down quarks in aneutron = 3 − n = 2
Therefore,a neutron can be represented as ‘udd’.
(a) Consider anarbitrary electrostatic field configuration. A small test charge is placed at anull point (i.e., where E =0) of the configuration. Show that the equilibrium of the test charge isnecessarily unstable.
(b) Verify this resultfor the simple configuration of two charges of the same magnitude and signplaced a certain distance apart.
Answer - 32 : -
(a) Let the equilibrium of thetest charge be stable. If a test charge is in equilibrium and displaced fromits position in any direction, then it experiences a restoring force towards anull point, where the electric field is zero. All the field lines near the nullpoint are directed inwards towards the null point. There is a net inward fluxof electric field through a closed surface around the null point. According toGauss’s law, the flux of electric field through a surface, which is notenclosing any charge, is zero. Hence, the equilibrium of the test charge can bestable.
(b) Two charges of samemagnitude and same sign are placed at a certain distance. The mid-point of thejoining line of the charges is the null point. When a test charged is displacedalong the line, it experiences a restoring force. If it is displaced normal tothe joining line, then the net force takes it away from the null point. Hence,the charge is unstable because stability of equilibrium requires restoringforce in all directions.
Question - 33 : - A particle of mass m andcharge (−q) enters the region between the two charged plates initiallymoving along x-axis with speed vx (like particle 1in Fig. 1.33). The length of plate is L and an uniformelectric field E is maintained between the plates. Show thatthe vertical deflection of the particle at the far edge of the plate is qEL2/(2m
).
Answer - 33 : -
Charge on a particle ofmass m = − q
Velocity of the particle= vx
Length of the plates = L
Magnitude of the uniformelectric field between the plates = E
Mechanical force, F =Mass (m) × Acceleration (a)
Therefore, acceleration, 
Time taken by the particleto cross the field of length L is given by,
t
In the vertical direction,initial velocity, u = 0
According to the thirdequation of motion, vertical deflection s of the particle canbe obtained as,
Hence, vertical deflectionof the particle at the far edge of the plate is
. This is similar to the motion of horizontalprojectiles under gravity.
Question - 34 : - Suppose that the particlein Exercise in 1.33 is an electron projected with velocity vx=2.0 × 106 m s−1. If E between theplates separated by 0.5 cm is 9.1 × 102 N/C, where will theelectron strike the upper plate? (| e | =1.6 × 10−19 C, me =9.1 × 10−31 kg.)
Answer - 34 : -
Velocity of the particle, vx =2.0 × 106 m/s
Separation of the twoplates, d = 0.5 cm = 0.005 m
Electric field between thetwo plates, E = 9.1 × 102 N/C
Charge on anelectron, q = 1.6 × 10−19 C
Mass of an electron, me =9.1 × 10−31 kg
Let the electron strikethe upper plate at the end of plate L, when deflection is s.
Therefore,Therefore, the electronwill strike the upper plate after travelling 1.6 cm.