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
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 less