Ans (a). Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the conductor, q = 6 x 10-7 C Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it. Ans (b). Electric fiRead more
Ans (a).
Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 6 x 10-7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
Ans (b).
Electric field E just outside the conductor is given by the relation,
E=1/4πε0 x q/r²
Where, ε0 = Permittivity of free space and = 9 x 109 Nm2C⁻²
Therefore, E = (9x 109 x 1.6 x 10-7)/(o.12)²
=10⁵ NC⁻¹
Therefore, the electric field just outside the sphere is 105 NC-1.
Ans (c).
Electric field at a point 18 m from the centre of the sphere = Ei
Distance of the point from the centre, d = 18 cm = 0.18 m
1 q 9 x 109 x 1.6 x 10-7
E=1/4πε0 x q/d²
= (9x 109 x 1.6 x 10-7)/(1.8 x 10⁻² )²
= 4.4 x 10⁴ NC-1
Therefore ,the electric field at a point 18cm from the centre of the sphere is 4.4 x 10⁴ NC-1
Velocity of the particle, Vx = 2.0 x 106 m/s Separation of the two plates, d = 0.5 cm = 0.005 m Electric field between the two plates, E = 9.1 x 102 N/C Charge on an electron, q = 1.6 x 10-19 C Mass of an electron, me= 9.1 x 10-31 kg Let the electron strike the upper plate at the end of plate L, wheRead more
Velocity of the particle, Vx = 2.0 x 106 m/s
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E = 9.1 x 102 N/C
Charge on an electron, q = 1.6 x 10–19 C
Mass of an electron, me= 9.1 x 10-31 kg
Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,
s= (qEL2)/(2mV2x)
=>L = √ (2smV2x)/qE
= √ 2 x 0.005 x 9.1 x 10-31)/(1.6 x 10–19 x 9.1 x 102
= √ 0.00025 = 0.016 m=1.6 cm
Therefore ,the electron will strike the upper plate after travelling 1.6cm.
Charge on a particle of mass m = — q Velocity of the particle = vx Length of the plates = L Magnitude of the uniform electric field between the plates = E Mechanical force, F = Mass (m) x Acceleration (a) =>a = F/m a = qE/m .................. (1) [as electric force, F = qE] Time tRead more
Charge on a particle of mass m = — q
Velocity of the particle = vx
Length of the plates = L
Magnitude of the uniform electric field between the plates = E
Mechanical force, F = Mass (m) x Acceleration (a)
=>a = F/m
a = qE/m ……………… (1) [as electric force, F = qE]
Time taken by the particle to cross the field of length L is given by,
t=(Length of the plate)/(Velocity of the particle)
t=L/Vx——————–(2)
In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical deflection s of the particle can be obtained as.
s = ut + at2/2
=> 0 + 1/2 (qE/m) (L/Vx)2 [From (1) and (2)]
=> s= (qEL2)/(2mV2x)
Hence, vertical deflection of the particle at the far edge of the plate is (qEL2)/(2mV2x).
This is similar to the motion of horizontal projectiles under gravity.
Ans (a). Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards tRead more
Ans (a).
Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
Ans (b).
Two charges of same magnitude and same sign are placed at a certain distance. The midpoint of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of (+2/3) e. Charge due to n up quarks =(2/3 x e)n Number of down quarks in a proton = 3 - n Each down quark has a charge of = -1/3 x e Charge due to (3 - n) down quarks = (-1/3 x e) (3 — n) Total charge on a protRead more
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of (+2/3) e.
Charge due to n up quarks =(2/3 x e)n
Number of down quarks in a proton = 3 – n
Each down quark has a charge of = -1/3 x e
Charge due to (3 – n) down quarks = (-1/3 x e) (3 — n)
Total charge on a proton = + e
Therefore e = (2/3 x e) n + (-1/3 x e) (3 — n)
=> e = 2ne/3 – e + ne/3
=> 2 e = ne => n = 2
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3- n = 3 — 2 = 1 .
Therefore, a proton can be represented as ‘uud’.
A neutron also has three quarks. Let there be n up quarks in a neutron.
Charge on a neutron due to n up quarks = (+2/3 e )n
Number of down quarks is 3-n ,and each having a charge of -1/3 e.
Charge on a neutron due to (3-n) down quarks = (-1/3e) (3-n)
A spherical conductor of radius 12 cm has a charge of 1.6 × 10^⁻7 C distributed uniformly on its surface. What is the electric field (a) inside the sphere (b) just outside the sphere (c) at a point 18 cm from the centre of the sphere?
Ans (a). Radius of the spherical conductor, r = 12 cm = 0.12 m Charge is uniformly distributed over the conductor, q = 6 x 10-7 C Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it. Ans (b). Electric fiRead more
Ans (a).
Radius of the spherical conductor, r = 12 cm = 0.12 m
Charge is uniformly distributed over the conductor, q = 6 x 10-7 C
Electric field inside a spherical conductor is zero. This is because if there is field inside the conductor, then charges will move to neutralize it.
Ans (b).
Electric field E just outside the conductor is given by the relation,
E=1/4πε0 x q/r²
Where, ε0 = Permittivity of free space and = 9 x 109 Nm2C⁻²
Therefore, E = (9x 109 x 1.6 x 10-7)/(o.12)²
=10⁵ NC⁻¹
Therefore, the electric field just outside the sphere is 105 NC-1.
Ans (c).
Electric field at a point 18 m from the centre of the sphere = Ei
Distance of the point from the centre, d = 18 cm = 0.18 m
1 q 9 x 109 x 1.6 x 10-7
E=1/4πε0 x q/d²
= (9x 109 x 1.6 x 10-7)/(1.8 x 10⁻² )²
= 4.4 x 10⁴ NC-1
Therefore ,the electric field at a point 18cm from the centre of the sphere is 4.4 x 10⁴ NC-1
See lessSuppose that the particle in Exercise in 1.33 is an electron projected with velocity vₓ = 2.0 × 10⁶ m s⁻¹. If E between the plates separated by 0.5 cm is 9.1 × 10² N/C, where will the electron strike the upper plate? (|e|=1.6 × 10⁻¹⁹ C, mₑ = 9.1 × 10 ⁻³¹kg.)
Velocity of the particle, Vx = 2.0 x 106 m/s Separation of the two plates, d = 0.5 cm = 0.005 m Electric field between the two plates, E = 9.1 x 102 N/C Charge on an electron, q = 1.6 x 10-19 C Mass of an electron, me= 9.1 x 10-31 kg Let the electron strike the upper plate at the end of plate L, wheRead more
Velocity of the particle, Vx = 2.0 x 106 m/s
Separation of the two plates, d = 0.5 cm = 0.005 m
Electric field between the two plates, E = 9.1 x 102 N/C
Charge on an electron, q = 1.6 x 10–19 C
Mass of an electron, me= 9.1 x 10-31 kg
Let the electron strike the upper plate at the end of plate L, when deflection is s. Therefore,
s= (qEL2)/(2mV2x)
=>L = √ (2smV2x)/qE
= √ 2 x 0.005 x 9.1 x 10-31)/(1.6 x 10–19 x 9.1 x 102
= √ 0.00025 = 0.016 m=1.6 cm
Therefore ,the electron will strike the upper plate after travelling 1.6cm.
See lessA particle of mass m and charge (–q) enters the region between the two charged plates initially moving along x-axis with speed vₓ (like particle 1 in Fig. 1.33). The length of plate is L and an uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL² /(2m vₓ² ). Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10 of Class XI Textbook of Physics.
Charge on a particle of mass m = — q Velocity of the particle = vx Length of the plates = L Magnitude of the uniform electric field between the plates = E Mechanical force, F = Mass (m) x Acceleration (a) =>a = F/m a = qE/m .................. (1) [as electric force, F = qE] Time tRead more
Charge on a particle of mass m = — q
Velocity of the particle = vx
Length of the plates = L
Magnitude of the uniform electric field between the plates = E
Mechanical force, F = Mass (m) x Acceleration (a)
=>a = F/m
a = qE/m ……………… (1) [as electric force, F = qE]
Time taken by the particle to cross the field of length L is given by,
t=(Length of the plate)/(Velocity of the particle)
t=L/Vx——————–(2)
In the vertical direction, initial velocity, u = 0
According to the third equation of motion, vertical deflection s of the particle can be obtained as.
s = ut + at2/2
=> 0 + 1/2 (qE/m) (L/Vx)2 [From (1) and (2)]
=> s= (qEL2)/(2mV2x)
Hence, vertical deflection of the particle at the far edge of the plate is (qEL2)/(2mV2x).
This is similar to the motion of horizontal projectiles under gravity.
See less2 (a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable. (b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Ans (a). Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards tRead more
Ans (a).
Let the equilibrium of the test charge be stable. If a test charge is in equilibrium and displaced from its position in any direction, then it experiences a restoring force towards a null point, where the electric field is zero. All the field lines near the null point are directed inwards towards the null point. There is a net inward flux of electric field through a closed surface around the null point. According to Gauss’s law, the flux of electric field through a surface, which is not enclosing any charge, is zero. Hence, the equilibrium of the test charge can be stable.
Ans (b).
Two charges of same magnitude and same sign are placed at a certain distance. The midpoint of the joining line of the charges is the null point. When a test charged is displaced along the line, it experiences a restoring force. If it is displaced normal to the joining line, then the net force takes it away from the null point. Hence, the charge is unstable because stability of equilibrium requires restoring force in all directions.
See lessIt is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks. A proton and a neutron 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 up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of (+2/3) e. Charge due to n up quarks =(2/3 x e)n Number of down quarks in a proton = 3 - n Each down quark has a charge of = -1/3 x e Charge due to (3 - n) down quarks = (-1/3 x e) (3 — n) Total charge on a protRead more
A proton has three quarks. Let there be n up quarks in a proton, each having a charge of (+2/3) e.
Charge due to n up quarks =(2/3 x e)n
Number of down quarks in a proton = 3 – n
Each down quark has a charge of = -1/3 x e
Charge due to (3 – n) down quarks = (-1/3 x e) (3 — n)
Total charge on a proton = + e
Therefore e = (2/3 x e) n + (-1/3 x e) (3 — n)
=> e = 2ne/3 – e + ne/3
=> 2 e = ne => n = 2
Number of up quarks in a proton, n = 2
Number of down quarks in a proton = 3- n = 3 — 2 = 1 .
Therefore, a proton can be represented as ‘uud’.
A neutron also has three quarks. Let there be n up quarks in a neutron.
Charge on a neutron due to n up quarks = (+2/3 e )n
Number of down quarks is 3-n ,and each having a charge of -1/3 e.
Charge on a neutron due to (3-n) down quarks = (-1/3e) (3-n)
Total charge on a neutron =0
Therefore 0=(2/3 x e) n + (-1/3 x e) (3-n)
=> 0 = 2ne/3 – e + ne/3
=> e= ne =>n=1
Number of up quarks in a neutron ,n=1
Number of down quarks in a neutron=3-n =2
Therefore, a neutron can be represented as ‘udd’.
See less