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1 | 890-893 | Choose
any convenient origin O and let the position vector of any point in the
charge distribution be r The charge density r may vary from point to
point, i e , it is a function of r |
1 | 891-894 | The charge density r may vary from point to
point, i e , it is a function of r Divide the charge distribution into small
volume elements of size DV |
1 | 892-895 | e , it is a function of r Divide the charge distribution into small
volume elements of size DV The charge in a volume element DV is rDV |
1 | 893-896 | , it is a function of r Divide the charge distribution into small
volume elements of size DV The charge in a volume element DV is rDV Now, consider any general point P (inside or outside the distribution)
with position vector R (Fig |
1 | 894-897 | Divide the charge distribution into small
volume elements of size DV The charge in a volume element DV is rDV Now, consider any general point P (inside or outside the distribution)
with position vector R (Fig 1 |
1 | 895-898 | The charge in a volume element DV is rDV Now, consider any general point P (inside or outside the distribution)
with position vector R (Fig 1 21) |
1 | 896-899 | Now, consider any general point P (inside or outside the distribution)
with position vector R (Fig 1 21) Electric field due to the charge rDV is
given by Coulomb’s law:
2
0
1
ˆ
4
V
'
r'
ρ
ε
∆
∆
=
π
E
r
(1 |
1 | 897-900 | 1 21) Electric field due to the charge rDV is
given by Coulomb’s law:
2
0
1
ˆ
4
V
'
r'
ρ
ε
∆
∆
=
π
E
r
(1 26)
where r¢ is the distance between the charge element and P, and ˆr¢ is a
unit vector in the direction from the charge element to P |
1 | 898-901 | 21) Electric field due to the charge rDV is
given by Coulomb’s law:
2
0
1
ˆ
4
V
'
r'
ρ
ε
∆
∆
=
π
E
r
(1 26)
where r¢ is the distance between the charge element and P, and ˆr¢ is a
unit vector in the direction from the charge element to P By the
superposition principle, the total electric field due to the charge
distribution is obtained by summing over electric fields due to different
volume elements:
2
0
1
ˆ
4
all
V
V
'
r'
ρ
ε
∆
∆
≅
Σ
π
E
r
(1 |
1 | 899-902 | Electric field due to the charge rDV is
given by Coulomb’s law:
2
0
1
ˆ
4
V
'
r'
ρ
ε
∆
∆
=
π
E
r
(1 26)
where r¢ is the distance between the charge element and P, and ˆr¢ is a
unit vector in the direction from the charge element to P By the
superposition principle, the total electric field due to the charge
distribution is obtained by summing over electric fields due to different
volume elements:
2
0
1
ˆ
4
all
V
V
'
r'
ρ
ε
∆
∆
≅
Σ
π
E
r
(1 27)
Note that r, r¢, ˆ′r all can vary from point to point |
1 | 900-903 | 26)
where r¢ is the distance between the charge element and P, and ˆr¢ is a
unit vector in the direction from the charge element to P By the
superposition principle, the total electric field due to the charge
distribution is obtained by summing over electric fields due to different
volume elements:
2
0
1
ˆ
4
all
V
V
'
r'
ρ
ε
∆
∆
≅
Σ
π
E
r
(1 27)
Note that r, r¢, ˆ′r all can vary from point to point In a strict
mathematical method, we should let DV®0 and the sum then becomes
an integral; but we omit that discussion here, for simplicity |
1 | 901-904 | By the
superposition principle, the total electric field due to the charge
distribution is obtained by summing over electric fields due to different
volume elements:
2
0
1
ˆ
4
all
V
V
'
r'
ρ
ε
∆
∆
≅
Σ
π
E
r
(1 27)
Note that r, r¢, ˆ′r all can vary from point to point In a strict
mathematical method, we should let DV®0 and the sum then becomes
an integral; but we omit that discussion here, for simplicity In short,
using Coulomb’s law and the superposition principle, electric field can
be determined for any charge distribution, discrete or continuous or part
discrete and part continuous |
1 | 902-905 | 27)
Note that r, r¢, ˆ′r all can vary from point to point In a strict
mathematical method, we should let DV®0 and the sum then becomes
an integral; but we omit that discussion here, for simplicity In short,
using Coulomb’s law and the superposition principle, electric field can
be determined for any charge distribution, discrete or continuous or part
discrete and part continuous 1 |
1 | 903-906 | In a strict
mathematical method, we should let DV®0 and the sum then becomes
an integral; but we omit that discussion here, for simplicity In short,
using Coulomb’s law and the superposition principle, electric field can
be determined for any charge distribution, discrete or continuous or part
discrete and part continuous 1 13 GAUSS’S LAW
As a simple application of the notion of electric flux, let us consider the
total flux through a sphere of radius r, which encloses a point charge q
at its centre |
1 | 904-907 | In short,
using Coulomb’s law and the superposition principle, electric field can
be determined for any charge distribution, discrete or continuous or part
discrete and part continuous 1 13 GAUSS’S LAW
As a simple application of the notion of electric flux, let us consider the
total flux through a sphere of radius r, which encloses a point charge q
at its centre Divide the sphere into small area elements, as shown in
Fig |
1 | 905-908 | 1 13 GAUSS’S LAW
As a simple application of the notion of electric flux, let us consider the
total flux through a sphere of radius r, which encloses a point charge q
at its centre Divide the sphere into small area elements, as shown in
Fig 1 |
1 | 906-909 | 13 GAUSS’S LAW
As a simple application of the notion of electric flux, let us consider the
total flux through a sphere of radius r, which encloses a point charge q
at its centre Divide the sphere into small area elements, as shown in
Fig 1 22 |
1 | 907-910 | Divide the sphere into small area elements, as shown in
Fig 1 22 The flux through an area element DS is
2
0
ˆ
4
q
r
φ
ε
∆
=
∆
=
∆
π
E
S
r
S
i
i
(1 |
1 | 908-911 | 1 22 The flux through an area element DS is
2
0
ˆ
4
q
r
φ
ε
∆
=
∆
=
∆
π
E
S
r
S
i
i
(1 28)
where we have used Coulomb’s law for the electric field due to a single
charge q |
1 | 909-912 | 22 The flux through an area element DS is
2
0
ˆ
4
q
r
φ
ε
∆
=
∆
=
∆
π
E
S
r
S
i
i
(1 28)
where we have used Coulomb’s law for the electric field due to a single
charge q The unit vector ˆr is along the radius vector from the centre to
the area element |
1 | 910-913 | The flux through an area element DS is
2
0
ˆ
4
q
r
φ
ε
∆
=
∆
=
∆
π
E
S
r
S
i
i
(1 28)
where we have used Coulomb’s law for the electric field due to a single
charge q The unit vector ˆr is along the radius vector from the centre to
the area element Now, since the normal to a sphere at every point is
along the radius vector at that point, the area element DS and ˆr have
the same direction |
1 | 911-914 | 28)
where we have used Coulomb’s law for the electric field due to a single
charge q The unit vector ˆr is along the radius vector from the centre to
the area element Now, since the normal to a sphere at every point is
along the radius vector at that point, the area element DS and ˆr have
the same direction Therefore,
2
0
4
q
S
r
φ
ε
∆
=
∆
π
(1 |
1 | 912-915 | The unit vector ˆr is along the radius vector from the centre to
the area element Now, since the normal to a sphere at every point is
along the radius vector at that point, the area element DS and ˆr have
the same direction Therefore,
2
0
4
q
S
r
φ
ε
∆
=
∆
π
(1 29)
since the magnitude of a unit vector is 1 |
1 | 913-916 | Now, since the normal to a sphere at every point is
along the radius vector at that point, the area element DS and ˆr have
the same direction Therefore,
2
0
4
q
S
r
φ
ε
∆
=
∆
π
(1 29)
since the magnitude of a unit vector is 1 The total flux through the sphere is obtained by adding up flux
through all the different area elements:
FIGURE 1 |
1 | 914-917 | Therefore,
2
0
4
q
S
r
φ
ε
∆
=
∆
π
(1 29)
since the magnitude of a unit vector is 1 The total flux through the sphere is obtained by adding up flux
through all the different area elements:
FIGURE 1 22 Flux
through a sphere
enclosing a point
charge q at its centre |
1 | 915-918 | 29)
since the magnitude of a unit vector is 1 The total flux through the sphere is obtained by adding up flux
through all the different area elements:
FIGURE 1 22 Flux
through a sphere
enclosing a point
charge q at its centre Rationalised 2023-24
30
Physics
2
0
4
all
S
q
S
r
φ
ε
∆
=
Σ
∆
π
Since each area element of the sphere is at the same
distance r from the charge,
2
2
0
4
4
all
S
o
q
q
S
S
r
r
φ
ε
ε
∆
=
Σ ∆
=
π
π
Now S, the total area of the sphere, equals 4pr 2 |
1 | 916-919 | The total flux through the sphere is obtained by adding up flux
through all the different area elements:
FIGURE 1 22 Flux
through a sphere
enclosing a point
charge q at its centre Rationalised 2023-24
30
Physics
2
0
4
all
S
q
S
r
φ
ε
∆
=
Σ
∆
π
Since each area element of the sphere is at the same
distance r from the charge,
2
2
0
4
4
all
S
o
q
q
S
S
r
r
φ
ε
ε
∆
=
Σ ∆
=
π
π
Now S, the total area of the sphere, equals 4pr 2 Thus,
2
2
0
0
4
4
q
q
r
r
φ
ε
ε
=
×
π
=
π
(1 |
1 | 917-920 | 22 Flux
through a sphere
enclosing a point
charge q at its centre Rationalised 2023-24
30
Physics
2
0
4
all
S
q
S
r
φ
ε
∆
=
Σ
∆
π
Since each area element of the sphere is at the same
distance r from the charge,
2
2
0
4
4
all
S
o
q
q
S
S
r
r
φ
ε
ε
∆
=
Σ ∆
=
π
π
Now S, the total area of the sphere, equals 4pr 2 Thus,
2
2
0
0
4
4
q
q
r
r
φ
ε
ε
=
×
π
=
π
(1 30)
Equation (1 |
1 | 918-921 | Rationalised 2023-24
30
Physics
2
0
4
all
S
q
S
r
φ
ε
∆
=
Σ
∆
π
Since each area element of the sphere is at the same
distance r from the charge,
2
2
0
4
4
all
S
o
q
q
S
S
r
r
φ
ε
ε
∆
=
Σ ∆
=
π
π
Now S, the total area of the sphere, equals 4pr 2 Thus,
2
2
0
0
4
4
q
q
r
r
φ
ε
ε
=
×
π
=
π
(1 30)
Equation (1 30) is a simple illustration of a general result of
electrostatics called Gauss’s law |
1 | 919-922 | Thus,
2
2
0
0
4
4
q
q
r
r
φ
ε
ε
=
×
π
=
π
(1 30)
Equation (1 30) is a simple illustration of a general result of
electrostatics called Gauss’s law We state Gauss’s law without proof:
Electric flux through a closed surface S
= q/e0
(1 |
1 | 920-923 | 30)
Equation (1 30) is a simple illustration of a general result of
electrostatics called Gauss’s law We state Gauss’s law without proof:
Electric flux through a closed surface S
= q/e0
(1 31)
q = total charge enclosed by S |
1 | 921-924 | 30) is a simple illustration of a general result of
electrostatics called Gauss’s law We state Gauss’s law without proof:
Electric flux through a closed surface S
= q/e0
(1 31)
q = total charge enclosed by S The law implies that the total electric flux through a closed surface is
zero if no charge is enclosed by the surface |
1 | 922-925 | We state Gauss’s law without proof:
Electric flux through a closed surface S
= q/e0
(1 31)
q = total charge enclosed by S The law implies that the total electric flux through a closed surface is
zero if no charge is enclosed by the surface We can see that explicitly in
the simple situation of Fig |
1 | 923-926 | 31)
q = total charge enclosed by S The law implies that the total electric flux through a closed surface is
zero if no charge is enclosed by the surface We can see that explicitly in
the simple situation of Fig 1 |
1 | 924-927 | The law implies that the total electric flux through a closed surface is
zero if no charge is enclosed by the surface We can see that explicitly in
the simple situation of Fig 1 23 |
1 | 925-928 | We can see that explicitly in
the simple situation of Fig 1 23 Here the electric field is uniform and we are considering a closed
cylindrical surface, with its axis parallel to the uniform field E |
1 | 926-929 | 1 23 Here the electric field is uniform and we are considering a closed
cylindrical surface, with its axis parallel to the uniform field E The total
flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent
the flux through the surfaces 1 and 2 (of circular cross-section) of the
cylinder and f3 is the flux through the curved cylindrical part of the
closed surface |
1 | 927-930 | 23 Here the electric field is uniform and we are considering a closed
cylindrical surface, with its axis parallel to the uniform field E The total
flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent
the flux through the surfaces 1 and 2 (of circular cross-section) of the
cylinder and f3 is the flux through the curved cylindrical part of the
closed surface Now the normal to the surface 3 at every point is
perpendicular to E, so by definition of flux, f3 = 0 |
1 | 928-931 | Here the electric field is uniform and we are considering a closed
cylindrical surface, with its axis parallel to the uniform field E The total
flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent
the flux through the surfaces 1 and 2 (of circular cross-section) of the
cylinder and f3 is the flux through the curved cylindrical part of the
closed surface Now the normal to the surface 3 at every point is
perpendicular to E, so by definition of flux, f3 = 0 Further, the outward
normal to 2 is along E while the outward normal to 1 is opposite to E |
1 | 929-932 | The total
flux f through the surface is f = f1 + f2 + f3, where f1 and f2 represent
the flux through the surfaces 1 and 2 (of circular cross-section) of the
cylinder and f3 is the flux through the curved cylindrical part of the
closed surface Now the normal to the surface 3 at every point is
perpendicular to E, so by definition of flux, f3 = 0 Further, the outward
normal to 2 is along E while the outward normal to 1 is opposite to E Therefore,
f1 = –E S1, f2 = +E S2
S1 = S2 = S
where S is the area of circular cross-section |
1 | 930-933 | Now the normal to the surface 3 at every point is
perpendicular to E, so by definition of flux, f3 = 0 Further, the outward
normal to 2 is along E while the outward normal to 1 is opposite to E Therefore,
f1 = –E S1, f2 = +E S2
S1 = S2 = S
where S is the area of circular cross-section Thus, the total flux is zero,
as expected by Gauss’s law |
1 | 931-934 | Further, the outward
normal to 2 is along E while the outward normal to 1 is opposite to E Therefore,
f1 = –E S1, f2 = +E S2
S1 = S2 = S
where S is the area of circular cross-section Thus, the total flux is zero,
as expected by Gauss’s law Thus, whenever you find that the net electric
flux through a closed surface is zero, we conclude that the total charge
contained in the closed surface is zero |
1 | 932-935 | Therefore,
f1 = –E S1, f2 = +E S2
S1 = S2 = S
where S is the area of circular cross-section Thus, the total flux is zero,
as expected by Gauss’s law Thus, whenever you find that the net electric
flux through a closed surface is zero, we conclude that the total charge
contained in the closed surface is zero The great significance of Gauss’s law Eq |
1 | 933-936 | Thus, the total flux is zero,
as expected by Gauss’s law Thus, whenever you find that the net electric
flux through a closed surface is zero, we conclude that the total charge
contained in the closed surface is zero The great significance of Gauss’s law Eq (1 |
1 | 934-937 | Thus, whenever you find that the net electric
flux through a closed surface is zero, we conclude that the total charge
contained in the closed surface is zero The great significance of Gauss’s law Eq (1 31), is that it is true in
general, and not only for the simple cases we have considered above |
1 | 935-938 | The great significance of Gauss’s law Eq (1 31), is that it is true in
general, and not only for the simple cases we have considered above Let
us note some important points regarding this law:
(i)
Gauss’s law is true for any closed surface, no matter what its shape
or size |
1 | 936-939 | (1 31), is that it is true in
general, and not only for the simple cases we have considered above Let
us note some important points regarding this law:
(i)
Gauss’s law is true for any closed surface, no matter what its shape
or size (ii) The term q on the right side of Gauss’s law, Eq |
1 | 937-940 | 31), is that it is true in
general, and not only for the simple cases we have considered above Let
us note some important points regarding this law:
(i)
Gauss’s law is true for any closed surface, no matter what its shape
or size (ii) The term q on the right side of Gauss’s law, Eq (1 |
1 | 938-941 | Let
us note some important points regarding this law:
(i)
Gauss’s law is true for any closed surface, no matter what its shape
or size (ii) The term q on the right side of Gauss’s law, Eq (1 31), includes the
sum of all charges enclosed by the surface |
1 | 939-942 | (ii) The term q on the right side of Gauss’s law, Eq (1 31), includes the
sum of all charges enclosed by the surface The charges may be located
anywhere inside the surface |
1 | 940-943 | (1 31), includes the
sum of all charges enclosed by the surface The charges may be located
anywhere inside the surface (iii) In the situation when the surface is so chosen that there are some
charges inside and some outside, the electric field [whose flux appears
on the left side of Eq |
1 | 941-944 | 31), includes the
sum of all charges enclosed by the surface The charges may be located
anywhere inside the surface (iii) In the situation when the surface is so chosen that there are some
charges inside and some outside, the electric field [whose flux appears
on the left side of Eq (1 |
1 | 942-945 | The charges may be located
anywhere inside the surface (iii) In the situation when the surface is so chosen that there are some
charges inside and some outside, the electric field [whose flux appears
on the left side of Eq (1 31)] is due to all the charges, both inside and
outside S |
1 | 943-946 | (iii) In the situation when the surface is so chosen that there are some
charges inside and some outside, the electric field [whose flux appears
on the left side of Eq (1 31)] is due to all the charges, both inside and
outside S The term q on the right side of Gauss’s law, however,
represents only the total charge inside S |
1 | 944-947 | (1 31)] is due to all the charges, both inside and
outside S The term q on the right side of Gauss’s law, however,
represents only the total charge inside S FIGURE 1 |
1 | 945-948 | 31)] is due to all the charges, both inside and
outside S The term q on the right side of Gauss’s law, however,
represents only the total charge inside S FIGURE 1 23 Calculation of the
flux of uniform electric field
through the surface of a cylinder |
1 | 946-949 | The term q on the right side of Gauss’s law, however,
represents only the total charge inside S FIGURE 1 23 Calculation of the
flux of uniform electric field
through the surface of a cylinder Rationalised 2023-24
Electric Charges
and Fields
31
EXAMPLE 1 |
1 | 947-950 | FIGURE 1 23 Calculation of the
flux of uniform electric field
through the surface of a cylinder Rationalised 2023-24
Electric Charges
and Fields
31
EXAMPLE 1 10
(iv) The surface that we choose for the application of Gauss’s law is called
the Gaussian surface |
1 | 948-951 | 23 Calculation of the
flux of uniform electric field
through the surface of a cylinder Rationalised 2023-24
Electric Charges
and Fields
31
EXAMPLE 1 10
(iv) The surface that we choose for the application of Gauss’s law is called
the Gaussian surface You may choose any Gaussian surface and
apply Gauss’s law |
1 | 949-952 | Rationalised 2023-24
Electric Charges
and Fields
31
EXAMPLE 1 10
(iv) The surface that we choose for the application of Gauss’s law is called
the Gaussian surface You may choose any Gaussian surface and
apply Gauss’s law However, take care not to let the Gaussian surface
pass through any discrete charge |
1 | 950-953 | 10
(iv) The surface that we choose for the application of Gauss’s law is called
the Gaussian surface You may choose any Gaussian surface and
apply Gauss’s law However, take care not to let the Gaussian surface
pass through any discrete charge This is because electric field due
to a system of discrete charges is not well defined at the location of
any charge |
1 | 951-954 | You may choose any Gaussian surface and
apply Gauss’s law However, take care not to let the Gaussian surface
pass through any discrete charge This is because electric field due
to a system of discrete charges is not well defined at the location of
any charge (As you go close to the charge, the field grows without
any bound |
1 | 952-955 | However, take care not to let the Gaussian surface
pass through any discrete charge This is because electric field due
to a system of discrete charges is not well defined at the location of
any charge (As you go close to the charge, the field grows without
any bound ) However, the Gaussian surface can pass through a
continuous charge distribution |
1 | 953-956 | This is because electric field due
to a system of discrete charges is not well defined at the location of
any charge (As you go close to the charge, the field grows without
any bound ) However, the Gaussian surface can pass through a
continuous charge distribution (v) Gauss’s law is often useful towards a much easier calculation of the
electrostatic field when the system has some symmetry |
1 | 954-957 | (As you go close to the charge, the field grows without
any bound ) However, the Gaussian surface can pass through a
continuous charge distribution (v) Gauss’s law is often useful towards a much easier calculation of the
electrostatic field when the system has some symmetry This is
facilitated by the choice of a suitable Gaussian surface |
1 | 955-958 | ) However, the Gaussian surface can pass through a
continuous charge distribution (v) Gauss’s law is often useful towards a much easier calculation of the
electrostatic field when the system has some symmetry This is
facilitated by the choice of a suitable Gaussian surface (vi) Finally, Gauss’s law is based on the inverse square dependence on
distance contained in the Coulomb’s law |
1 | 956-959 | (v) Gauss’s law is often useful towards a much easier calculation of the
electrostatic field when the system has some symmetry This is
facilitated by the choice of a suitable Gaussian surface (vi) Finally, Gauss’s law is based on the inverse square dependence on
distance contained in the Coulomb’s law Any violation of Gauss’s
law will indicate departure from the inverse square law |
1 | 957-960 | This is
facilitated by the choice of a suitable Gaussian surface (vi) Finally, Gauss’s law is based on the inverse square dependence on
distance contained in the Coulomb’s law Any violation of Gauss’s
law will indicate departure from the inverse square law Example 1 |
1 | 958-961 | (vi) Finally, Gauss’s law is based on the inverse square dependence on
distance contained in the Coulomb’s law Any violation of Gauss’s
law will indicate departure from the inverse square law Example 1 10 The electric field components in Fig |
1 | 959-962 | Any violation of Gauss’s
law will indicate departure from the inverse square law Example 1 10 The electric field components in Fig 1 |
1 | 960-963 | Example 1 10 The electric field components in Fig 1 24 are
Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2 |
1 | 961-964 | 10 The electric field components in Fig 1 24 are
Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2 Calculate (a) the
flux through the cube, and (b) the charge within the cube |
1 | 962-965 | 1 24 are
Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2 Calculate (a) the
flux through the cube, and (b) the charge within the cube Assume
that a = 0 |
1 | 963-966 | 24 are
Ex = ax1/2, Ey = Ez = 0, in which a = 800 N/C m1/2 Calculate (a) the
flux through the cube, and (b) the charge within the cube Assume
that a = 0 1 m |
1 | 964-967 | Calculate (a) the
flux through the cube, and (b) the charge within the cube Assume
that a = 0 1 m FIGURE 1 |
1 | 965-968 | Assume
that a = 0 1 m FIGURE 1 24
Solution
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and DS is
± p/2 |
1 | 966-969 | 1 m FIGURE 1 24
Solution
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and DS is
± p/2 Therefore, the flux f = E |
1 | 967-970 | FIGURE 1 24
Solution
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and DS is
± p/2 Therefore, the flux f = E DS is separately zero for each face
of the cube except the two shaded ones |
1 | 968-971 | 24
Solution
(a) Since the electric field has only an x component, for faces
perpendicular to x direction, the angle between E and DS is
± p/2 Therefore, the flux f = E DS is separately zero for each face
of the cube except the two shaded ones Now the magnitude of
the electric field at the left face is
EL = ax1/2 = aa1/2
(x = a at the left face) |
1 | 969-972 | Therefore, the flux f = E DS is separately zero for each face
of the cube except the two shaded ones Now the magnitude of
the electric field at the left face is
EL = ax1/2 = aa1/2
(x = a at the left face) The magnitude of electric field at the right face is
ER = a x1/2 = a (2a)1/2
(x = 2a at the right face) |
1 | 970-973 | DS is separately zero for each face
of the cube except the two shaded ones Now the magnitude of
the electric field at the left face is
EL = ax1/2 = aa1/2
(x = a at the left face) The magnitude of electric field at the right face is
ER = a x1/2 = a (2a)1/2
(x = 2a at the right face) The corresponding fluxes are
fL= EL |
1 | 971-974 | Now the magnitude of
the electric field at the left face is
EL = ax1/2 = aa1/2
(x = a at the left face) The magnitude of electric field at the right face is
ER = a x1/2 = a (2a)1/2
(x = 2a at the right face) The corresponding fluxes are
fL= EL DS =
ˆ
L
L
∆S
E
⋅n
=EL DS cosq = –EL DS, since q = 180°
= –ELa2
fR= ER |
1 | 972-975 | The magnitude of electric field at the right face is
ER = a x1/2 = a (2a)1/2
(x = 2a at the right face) The corresponding fluxes are
fL= EL DS =
ˆ
L
L
∆S
E
⋅n
=EL DS cosq = –EL DS, since q = 180°
= –ELa2
fR= ER DS = ER DS cosq = ER DS, since q = 0°
= ERa2
Net flux through the cube
Rationalised 2023-24
32
Physics
EXAMPLE 1 |
1 | 973-976 | The corresponding fluxes are
fL= EL DS =
ˆ
L
L
∆S
E
⋅n
=EL DS cosq = –EL DS, since q = 180°
= –ELa2
fR= ER DS = ER DS cosq = ER DS, since q = 0°
= ERa2
Net flux through the cube
Rationalised 2023-24
32
Physics
EXAMPLE 1 11
EXAMPLE 1 |
1 | 974-977 | DS =
ˆ
L
L
∆S
E
⋅n
=EL DS cosq = –EL DS, since q = 180°
= –ELa2
fR= ER DS = ER DS cosq = ER DS, since q = 0°
= ERa2
Net flux through the cube
Rationalised 2023-24
32
Physics
EXAMPLE 1 11
EXAMPLE 1 10
= fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2]
= aa5/2 (
)
2 – 1
= 800 (0 |
1 | 975-978 | DS = ER DS cosq = ER DS, since q = 0°
= ERa2
Net flux through the cube
Rationalised 2023-24
32
Physics
EXAMPLE 1 11
EXAMPLE 1 10
= fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2]
= aa5/2 (
)
2 – 1
= 800 (0 1)5/2 (
)
2 – 1
= 1 |
1 | 976-979 | 11
EXAMPLE 1 10
= fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2]
= aa5/2 (
)
2 – 1
= 800 (0 1)5/2 (
)
2 – 1
= 1 05 N m2 C–1
(b) We can use Gauss’s law to find the total charge q inside the cube |
1 | 977-980 | 10
= fR + fL = ERa2 – ELa2 = a2 (ER – EL) = aa2 [(2a)1/2 – a1/2]
= aa5/2 (
)
2 – 1
= 800 (0 1)5/2 (
)
2 – 1
= 1 05 N m2 C–1
(b) We can use Gauss’s law to find the total charge q inside the cube We have f = q/e0 or q = fe0 |
1 | 978-981 | 1)5/2 (
)
2 – 1
= 1 05 N m2 C–1
(b) We can use Gauss’s law to find the total charge q inside the cube We have f = q/e0 or q = fe0 Therefore,
q = 1 |
1 | 979-982 | 05 N m2 C–1
(b) We can use Gauss’s law to find the total charge q inside the cube We have f = q/e0 or q = fe0 Therefore,
q = 1 05 × 8 |
1 | 980-983 | We have f = q/e0 or q = fe0 Therefore,
q = 1 05 × 8 854 × 10–12 C = 9 |
1 | 981-984 | Therefore,
q = 1 05 × 8 854 × 10–12 C = 9 27 × 10–12 C |
1 | 982-985 | 05 × 8 854 × 10–12 C = 9 27 × 10–12 C Example 1 |
1 | 983-986 | 854 × 10–12 C = 9 27 × 10–12 C Example 1 11 An electric field is uniform, and in the positive x
direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x |
1 | 984-987 | 27 × 10–12 C Example 1 11 An electric field is uniform, and in the positive x
direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x It is given that E = 200 ˆi N/C
for x > 0 and E = –200 ˆi N/C for x < 0 |
1 | 985-988 | Example 1 11 An electric field is uniform, and in the positive x
direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x It is given that E = 200 ˆi N/C
for x > 0 and E = –200 ˆi N/C for x < 0 A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = –10 cm (Fig |
1 | 986-989 | 11 An electric field is uniform, and in the positive x
direction for positive x, and uniform with the same magnitude but in
the negative x direction for negative x It is given that E = 200 ˆi N/C
for x > 0 and E = –200 ˆi N/C for x < 0 A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = –10 cm (Fig 1 |
1 | 987-990 | It is given that E = 200 ˆi N/C
for x > 0 and E = –200 ˆi N/C for x < 0 A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = –10 cm (Fig 1 25) |
1 | 988-991 | A right circular cylinder of
length 20 cm and radius 5 cm has its centre at the origin and its axis
along the x-axis so that one face is at x = +10 cm and the other is at
x = –10 cm (Fig 1 25) (a) What is the net outward flux through each
flat face |
1 | 989-992 | 1 25) (a) What is the net outward flux through each
flat face (b) What is the flux through the side of the cylinder |
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