Ans (a). Since diameter of the sphere, d = 4 m, therefore Radius of the sphere, r = 1.2 m Surface charge density, σ =80.0 µC/m2 = 80 x 10-6 C/m2 Total charge on the surface of the sphere, Q = Charge density x Surface area = σ x 4 π r2 = 80 x 10-6 x 4 x 3.14 x (1.2)2 = 1.447 x 10-3 C Therefore, the cRead more
Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation, E= 1 /4πε0 x q/d² Where, q = Net charge = 1.5 x 103 N/C d = Distance from the centre of sphere = 20 cm = 0.2 m ε0 = Permittivity of free space and 1 /4πε0 = 9 x109Nm2C⁻2Read more
Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,
E= 1 /4πε0 x q/d²
Where, q = Net charge = 1.5 x 103 N/C
d = Distance from the centre of sphere = 20 cm = 0.2 m
ε0 = Permittivity of free space and
1 /4πε0 = 9 x109Nm2C⁻2
Therefore,
q = E(4πε0)d2 =
=(1.5 x 10³) (0.2)2 / (9 x 109)
6.67 x 10⁻9C =6.67 nC
Therefore, the net charge on the sphere is 6.67 nC.
Ans (a). Electric flux, Φ = —1.0 x 103 N m2/C and radius of the Gaussian surface, r = 10.0 cm Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux pasRead more
Ans (a).
Electric flux, Φ = —1.0 x 103 N m2/C and radius of the Gaussian surface, r = 10.0 cm Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., —103 N m2/C.
Ans (b).
Electric flux is given by the relation
Φ=q/ε0
Where ε0 = Permittivity of free space = 8.854 x 10–12 N–1C2 m-2
q = Net charge enclosed by the spherical surface = Φε0
= -1.0 x 103 x 8.854 x 10–12 =-8.854 x 10–9C = -8.854 nC
Therefore, the value of the point charge is -8.854 nC.
Net electric flux (ΦNet) through the cubic surface is given by ΦNet =q/ε0 Where, ε0 = Permittivity of free space = 8.854 x 10⁻12 N⁻1C2 m⁻2 q = Net charge contained inside the cube = 2.0 µC = 2 x 10⁻6 C Therefore ΦNet =(2 x 10⁻6)/8.854 x 10⁻12 =2.26 x 10⁵Nm2C⁻1 The net electric flux through the surfaRead more
Net electric flux (ΦNet) through the cubic surface is given by
ΦNet =q/ε0
Where,
ε0 = Permittivity of free space = 8.854 x 10⁻12 N⁻1C2 m⁻2
q = Net charge contained inside the cube = 2.0 µC = 2 x 10⁻6 C
Therefore ΦNet =(2 x 10⁻6)/8.854 x 10⁻12
=2.26 x 10⁵Nm2C⁻1
The net electric flux through the surface is 2.26 x 10⁵Nm2C⁻1
Refer Figure 1.34 Exercise 1.18 The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux through all its six faces. ΦTotal = q/ε0 Hence, electric flux through one face of the cube i.e., throRead more
Refer Figure 1.34 Exercise 1.18
The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux through all its six faces.
ΦTotal = q/ε0
Hence, electric flux through one face of the cube i.e., through the square is
Φ=ΦTotal /6 =1/6 x q/ε0
Where,
ε0 = Permittivity of free space = 8.854 x 10–12 N–1C2 m-2
q = 10 µ C = 10 x 10-6 C
Therefore Φ=1/6 x (10 x 10-6)/ 8.854 x 10-12
= 1.88 x 105 N m2 C-1
Therefore, electric flux through the square is 1.88 x 105 N m2 C-1
Ans (a). Net outward flux through the surface of the box, Φ = 8.0 x 103 N m2/C For a body containing net charge q, flux is given by the relation, Φ=q/ε0 Therefore charge q = Φε0 where ε0 = Permittivity of free space =8.854 x 10⁻12 N⁻¹C2 m⁻2 q = Φε0 = 8.854 x 10⁻12 x 8.0 x 103 C = 7.08 x 10⁻8 C = 0Read more
Ans (a).
Net outward flux through the surface of the box, Φ = 8.0 x 103 N m2/C
For a body containing net charge q, flux is given by the relation,
Φ=q/ε0 Therefore charge q = Φε0
where ε0 = Permittivity of free space =8.854 x 10⁻12 N⁻¹C2 m⁻2
q = Φε0
= 8.854 x 10⁻12 x 8.0 x 103 C = 7.08 x 10⁻8 C = 0.07 µC.
Therefore, the net charge inside the box is 0.07 µC.
(b) No
Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.
Refer Exercise 1.15 All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.
Refer Exercise 1.15
All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.
Ans (a). Electric field intensity, E = 3 x 103 î N/C Magnitude of electric field intensity, |E| = 3 x 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 is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane andRead more
Ans (a).
Electric field intensity, E = 3 x 103î N/C
Magnitude of electric field intensity, |E| = 3 x 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 is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, Θ = 0°
Flux (Φ ) through the plane is given by the relation,
Φ = |E| Acos Θ = 3 x 103 x 0.01 x cos 0° = 30 Nm²/C
Ans (b).
Plane makes an angle of 60° with the x – axis. Hence, Θ = 60°
Refer figure 1.33 in Exercise 1.14 Opposite charges attract each other and similar charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. IRead more
Refer figure 1.33 in Exercise 1.14
Opposite charges attract each other and similar charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged. The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.
In Exercise 1.12 the distance between the spheres, A and B was r = 0.5 m and the charge on each sphere, q = 6.5 x 10⁻7 C When sphere A is touched with an neutral (uncharged) sphere C, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, becomes q/2. WRead more
In Exercise 1.12 the distance between the spheres, A and B was r = 0.5 m and the charge on each sphere, q = 6.5 x 10⁻7 C
When sphere A is touched with an neutral (uncharged) sphere C, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, becomes q/2.
When sphere C with charge q/2 is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given by
1/2 x (q+q/2)=1/2 x 3q/2
=3q/4
Therefore the Force of repulsion between sphere A with charge q/2 and sphere B with charge 3q/4 will be
F= 1 /4πε0 x qAqB/r²
= 1 /4πε0 x (q/2)(3q/4)/r²
=1 /4πε0 x (3q²/8)/r²
=9 x 10⁹ x 3 x (6.5 x 10⁻7)²/(8 x (0.5)²
=5.703 x 10⁻³N
Therefore ,the force of attraction between the two spheres is 5.703 x 10⁻³N.
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?
Ans (a). Since diameter of the sphere, d = 4 m, therefore Radius of the sphere, r = 1.2 m Surface charge density, σ =80.0 µC/m2 = 80 x 10-6 C/m2 Total charge on the surface of the sphere, Q = Charge density x Surface area = σ x 4 π r2 = 80 x 10-6 x 4 x 3.14 x (1.2)2 = 1.447 x 10-3 C Therefore, the cRead more
Since diameter of the sphere, d = 4 m,
therefore Radius of the sphere, r = 1.2 m
Surface charge density, σ =80.0 µC/m2 = 80 x 10-6 C/m2
Total charge on the surface of the sphere,
Q = Charge density x Surface area = σ x 4 π r2 = 80 x 10-6 x 4 x 3.14 x (1.2)2 = 1.447 x 10-3 C Therefore, the charge on the sphere is 1.447 x 10-3 C.
Ans (b).
Total electric flux (ΦTotal ) leaving out the surface of a sphere containing net charge Q is given by the relation,
ΦTotal =Q/ε0
Where, ε0 = Permittivity of free space = 8.854 x 10-12 N-1C2 m-2
Q = 1.447 x 10–3C
Therefore ΦTotal =Q/ε0 = ( 1.447 x 10–3)/(8.854 x 10-12 )
= 1.63 x 108NC-1m2
Therefore, the total electric flux leaving the surface of the sphere is 1.63 x 108 N C-1 m2.
See lessA 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 × 10³N/C and points radially inward, what is the net charge on the sphere?
Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation, E= 1 /4πε0 x q/d² Where, q = Net charge = 1.5 x 103 N/C d = Distance from the centre of sphere = 20 cm = 0.2 m ε0 = Permittivity of free space and 1 /4πε0 = 9 x109Nm2C⁻2Read more
Electric field intensity (E) at a distance (d) from the centre of a sphere containing net charge q is given by the relation,
E= 1 /4πε0 x q/d²
Where, q = Net charge = 1.5 x 103 N/C
d = Distance from the centre of sphere = 20 cm = 0.2 m
ε0 = Permittivity of free space and
1 /4πε0 = 9 x109Nm2C⁻2
Therefore,
q = E(4πε0)d2 =
=(1.5 x 10³) (0.2)2 / (9 x 109)
6.67 x 10⁻9C =6.67 nC
Therefore, the net charge on the sphere is 6.67 nC.
See lessA point charge causes an electric flux of –1.0 × 10³Nm²/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?
Ans (a). Electric flux, Φ = —1.0 x 103 N m2/C and radius of the Gaussian surface, r = 10.0 cm Electric flux piercing out through a surface depends on the net charge enclosed inside a body. It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux pasRead more
Ans (a).
Electric flux is given by the relation
Φ=q/ε0
Where ε0 = Permittivity of free space = 8.854 x 10–12 N–1C2 m-2
q = Net charge enclosed by the spherical surface = Φε0
= -1.0 x 103 x 8.854 x 10–12 =-8.854 x 10–9C = -8.854 nC
Therefore, the value of the point charge is -8.854 nC.
See lessA 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?
Net electric flux (ΦNet) through the cubic surface is given by ΦNet =q/ε0 Where, ε0 = Permittivity of free space = 8.854 x 10⁻12 N⁻1C2 m⁻2 q = Net charge contained inside the cube = 2.0 µC = 2 x 10⁻6 C Therefore ΦNet =(2 x 10⁻6)/8.854 x 10⁻12 =2.26 x 10⁵Nm2C⁻1 The net electric flux through the surfaRead more
Net electric flux (ΦNet) through the cubic surface is given by
ΦNet =q/ε0
Where,
ε0 = Permittivity of free space = 8.854 x 10⁻12 N⁻1C2 m⁻2
q = Net charge contained inside the cube = 2.0 µC = 2 x 10⁻6 C
Therefore ΦNet =(2 x 10⁻6)/8.854 x 10⁻12
=2.26 x 10⁵Nm2C⁻1
The net electric flux through the surface is 2.26 x 10⁵Nm2C⁻1
See lessA 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.)
Refer Figure 1.34 Exercise 1.18 The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux through all its six faces. ΦTotal = q/ε0 Hence, electric flux through one face of the cube i.e., throRead more
Refer Figure 1.34 Exercise 1.18
The square can be considered as one face of a cube of edge 10 cm with a centre where charge q is placed. According to Gauss’s theorem for a cube, total electric flux through all its six faces.
ΦTotal = q/ε0
Hence, electric flux through one face of the cube i.e., through the square is
Φ=ΦTotal /6 =1/6 x q/ε0
Where,
ε0 = Permittivity of free space = 8.854 x 10–12 N–1C2 m-2
q = 10 µ C = 10 x 10-6 C
Therefore Φ=1/6 x (10 x 10-6)/ 8.854 x 10-12
= 1.88 x 105 N m2 C-1
Therefore, electric flux through the square is 1.88 x 105 N m2 C-1
See lessCareful 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?
Ans (a). Net outward flux through the surface of the box, Φ = 8.0 x 103 N m2/C For a body containing net charge q, flux is given by the relation, Φ=q/ε0 Therefore charge q = Φε0 where ε0 = Permittivity of free space =8.854 x 10⁻12 N⁻¹C2 m⁻2 q = Φε0 = 8.854 x 10⁻12 x 8.0 x 103 C = 7.08 x 10⁻8 C = 0Read more
Ans (a).
Net outward flux through the surface of the box, Φ = 8.0 x 103 N m2/C
For a body containing net charge q, flux is given by the relation,
Φ=q/ε0 Therefore charge q = Φε0
where ε0 = Permittivity of free space =8.854 x 10⁻12 N⁻¹C2 m⁻2
q = Φε0
= 8.854 x 10⁻12 x 8.0 x 103 C = 7.08 x 10⁻8 C = 0.07 µC.
Therefore, the net charge inside the box is 0.07 µC.
(b) No
Net flux piercing out through a body depends on the net charge contained in the body. If net flux is zero, then it can be inferred that net charge inside the body is zero. The body may have equal amount of positive and negative charges.
See lessWhat 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?
Refer Exercise 1.15 All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.
Refer Exercise 1.15
All the faces of a cube are parallel to the coordinate axes. Therefore, the number of field lines entering the cube is equal to the number of field lines piercing out of the cube. As a result, net flux through the cube is zero.
See lessConsider a uniform electric field E = 3 × 10³ î 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?
Ans (a). Electric field intensity, E = 3 x 103 î N/C Magnitude of electric field intensity, |E| = 3 x 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 is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane andRead more
Ans (a).
Electric field intensity, E = 3 x 103 î N/C
Magnitude of electric field intensity, |E| = 3 x 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 is parallel to the y-z plane. Hence, angle between the unit vector normal to the plane and electric field, Θ = 0°
Flux (Φ ) through the plane is given by the relation,
Φ = |E| Acos Θ = 3 x 103 x 0.01 x cos 0° = 30 Nm²/C
Ans (b).
Plane makes an angle of 60° with the x – axis. Hence, Θ = 60°
Flux, Φ = |E|Aco5 6 = 3 x 103 x 0.01 x cos 60°
= 30 x 1/2 = 15 Nm2/C
See lessFigure 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?
Refer figure 1.33 in Exercise 1.14 Opposite charges attract each other and similar charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. IRead more
Refer figure 1.33 in Exercise 1.14
See lessOpposite charges attract each other and similar charges repel each other. It can be observed that particles 1 and 2 both move towards the positively charged plate and repel away from the negatively charged plate. Hence, these two particles are negatively charged. It can also be observed that particle 3 moves towards the negatively charged plate and repels away from the positively charged plate. Hence, particle 3 is positively charged.
The charge to mass ratio (emf) is directly proportional to the displacement or amount of deflection for a given velocity. Since the deflection of particle 3 is the maximum, it has the highest charge to mass ratio.
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?
In Exercise 1.12 the distance between the spheres, A and B was r = 0.5 m and the charge on each sphere, q = 6.5 x 10⁻7 C When sphere A is touched with an neutral (uncharged) sphere C, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, becomes q/2. WRead more
In Exercise 1.12 the distance between the spheres, A and B was r = 0.5 m and the charge on each sphere, q = 6.5 x 10⁻7 C
When sphere A is touched with an neutral (uncharged) sphere C, q/2 amount of charge from A will transfer to sphere C. Hence, charge on each of the spheres, A and C, becomes q/2.
When sphere C with charge q/2 is brought in contact with sphere B with charge q, total charges on the system will divide into two equal halves given by
1/2 x (q+q/2)=1/2 x 3q/2
=3q/4
Therefore the Force of repulsion between sphere A with charge q/2 and sphere B with charge 3q/4 will be
F= 1 /4πε0 x qAqB/r²
= 1 /4πε0 x (q/2)(3q/4)/r²
=1 /4πε0 x (3q²/8)/r²
=9 x 10⁹ x 3 x (6.5 x 10⁻7)²/(8 x (0.5)²
=5.703 x 10⁻³N
Therefore ,the force of attraction between the two spheres is 5.703 x 10⁻³N.
See less