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1 | 1790-1793 | Solution
(a)
12
9
1
2
0
1
7
( 2)
10
9
10
4
0 18
q q
U
r
ε
−
× −
×
=
=
×
×
π
= –0 7 J (b) W = U2 – U1 = 0 – U = 0 – (–0 |
1 | 1791-1794 | 18
q q
U
r
ε
−
× −
×
=
=
×
×
π
= –0 7 J (b) W = U2 – U1 = 0 – U = 0 – (–0 7) = 0 |
1 | 1792-1795 | 7 J (b) W = U2 – U1 = 0 – U = 0 – (–0 7) = 0 7 J |
1 | 1793-1796 | (b) W = U2 – U1 = 0 – U = 0 – (–0 7) = 0 7 J (c) The mutual interaction energy of the two charges remains
unchanged |
1 | 1794-1797 | 7) = 0 7 J (c) The mutual interaction energy of the two charges remains
unchanged In addition, there is the energy of interaction of the
two charges with the external electric field |
1 | 1795-1798 | 7 J (c) The mutual interaction energy of the two charges remains
unchanged In addition, there is the energy of interaction of the
two charges with the external electric field We find,
( )
(
)
1
1
2
2
7 C
2 C
0 |
1 | 1796-1799 | (c) The mutual interaction energy of the two charges remains
unchanged In addition, there is the energy of interaction of the
two charges with the external electric field We find,
( )
(
)
1
1
2
2
7 C
2 C
0 09m
0 |
1 | 1797-1800 | In addition, there is the energy of interaction of the
two charges with the external electric field We find,
( )
(
)
1
1
2
2
7 C
2 C
0 09m
0 09m
q V
q V
A
A
µ
− µ
+
=
+
r
r
and the net electrostatic energy is
( )
(
)
1
2
1
1
2
2
0 12
7 C
2 C
0 |
1 | 1798-1801 | We find,
( )
(
)
1
1
2
2
7 C
2 C
0 09m
0 09m
q V
q V
A
A
µ
− µ
+
=
+
r
r
and the net electrostatic energy is
( )
(
)
1
2
1
1
2
2
0 12
7 C
2 C
0 7 J
4
0 |
1 | 1799-1802 | 09m
0 09m
q V
q V
A
A
µ
− µ
+
=
+
r
r
and the net electrostatic energy is
( )
(
)
1
2
1
1
2
2
0 12
7 C
2 C
0 7 J
4
0 09 m
0 |
1 | 1800-1803 | 09m
q V
q V
A
A
µ
− µ
+
=
+
r
r
and the net electrostatic energy is
( )
(
)
1
2
1
1
2
2
0 12
7 C
2 C
0 7 J
4
0 09 m
0 09 m
q q
q V
q V
A
A
r
ε
µ
− µ
+
+
=
+
−
π
r
r
70
20
0 |
1 | 1801-1804 | 7 J
4
0 09 m
0 09 m
q q
q V
q V
A
A
r
ε
µ
− µ
+
+
=
+
−
π
r
r
70
20
0 7
49 |
1 | 1802-1805 | 09 m
0 09 m
q q
q V
q V
A
A
r
ε
µ
− µ
+
+
=
+
−
π
r
r
70
20
0 7
49 3 J
=
−
−
=
2 |
1 | 1803-1806 | 09 m
q q
q V
q V
A
A
r
ε
µ
− µ
+
+
=
+
−
π
r
r
70
20
0 7
49 3 J
=
−
−
=
2 8 |
1 | 1804-1807 | 7
49 3 J
=
−
−
=
2 8 3 Potential energy of a dipole in an external field
Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform
electric field E, as shown in Fig |
1 | 1805-1808 | 3 J
=
−
−
=
2 8 3 Potential energy of a dipole in an external field
Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform
electric field E, as shown in Fig 2 |
1 | 1806-1809 | 8 3 Potential energy of a dipole in an external field
Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform
electric field E, as shown in Fig 2 16 |
1 | 1807-1810 | 3 Potential energy of a dipole in an external field
Consider a dipole with charges q1 = +q and q2 = –q placed in a uniform
electric field E, as shown in Fig 2 16 As seen in the last chapter, in a uniform electric field,
the dipole experiences no net force; but experiences a
torque t t t t t given by
t =
t =
t =
t =
t = p × E
(2 |
1 | 1808-1811 | 2 16 As seen in the last chapter, in a uniform electric field,
the dipole experiences no net force; but experiences a
torque t t t t t given by
t =
t =
t =
t =
t = p × E
(2 30)
which will tend to rotate it (unless p is parallel or
antiparallel to E) |
1 | 1809-1812 | 16 As seen in the last chapter, in a uniform electric field,
the dipole experiences no net force; but experiences a
torque t t t t t given by
t =
t =
t =
t =
t = p × E
(2 30)
which will tend to rotate it (unless p is parallel or
antiparallel to E) Suppose an external torque tttttext is
applied in such a manner that it just neutralises this
torque and rotates it in the plane of paper from angle q0
to angle q1 at an infinitesimal angular speed and without
angular acceleration |
1 | 1810-1813 | As seen in the last chapter, in a uniform electric field,
the dipole experiences no net force; but experiences a
torque t t t t t given by
t =
t =
t =
t =
t = p × E
(2 30)
which will tend to rotate it (unless p is parallel or
antiparallel to E) Suppose an external torque tttttext is
applied in such a manner that it just neutralises this
torque and rotates it in the plane of paper from angle q0
to angle q1 at an infinitesimal angular speed and without
angular acceleration The amount of work done by the
external torque will be given by
(
)
cos
cos
pE
θ
θ
0
1
=
−
(2 |
1 | 1811-1814 | 30)
which will tend to rotate it (unless p is parallel or
antiparallel to E) Suppose an external torque tttttext is
applied in such a manner that it just neutralises this
torque and rotates it in the plane of paper from angle q0
to angle q1 at an infinitesimal angular speed and without
angular acceleration The amount of work done by the
external torque will be given by
(
)
cos
cos
pE
θ
θ
0
1
=
−
(2 31)
This work is stored as the potential energy of the system |
1 | 1812-1815 | Suppose an external torque tttttext is
applied in such a manner that it just neutralises this
torque and rotates it in the plane of paper from angle q0
to angle q1 at an infinitesimal angular speed and without
angular acceleration The amount of work done by the
external torque will be given by
(
)
cos
cos
pE
θ
θ
0
1
=
−
(2 31)
This work is stored as the potential energy of the system We can
then associate potential energy U(q) with an inclination q of the dipole |
1 | 1813-1816 | The amount of work done by the
external torque will be given by
(
)
cos
cos
pE
θ
θ
0
1
=
−
(2 31)
This work is stored as the potential energy of the system We can
then associate potential energy U(q) with an inclination q of the dipole Similar to other potential energies, there is a freedom in choosing the
angle where the potential energy U is taken to be zero |
1 | 1814-1817 | 31)
This work is stored as the potential energy of the system We can
then associate potential energy U(q) with an inclination q of the dipole Similar to other potential energies, there is a freedom in choosing the
angle where the potential energy U is taken to be zero A natural choice
is to take q0 = p / 2 |
1 | 1815-1818 | We can
then associate potential energy U(q) with an inclination q of the dipole Similar to other potential energies, there is a freedom in choosing the
angle where the potential energy U is taken to be zero A natural choice
is to take q0 = p / 2 (An explanation for it is provided towards the end of
discussion |
1 | 1816-1819 | Similar to other potential energies, there is a freedom in choosing the
angle where the potential energy U is taken to be zero A natural choice
is to take q0 = p / 2 (An explanation for it is provided towards the end of
discussion ) We can then write,
(2 |
1 | 1817-1820 | A natural choice
is to take q0 = p / 2 (An explanation for it is provided towards the end of
discussion ) We can then write,
(2 32)
FIGURE 2 |
1 | 1818-1821 | (An explanation for it is provided towards the end of
discussion ) We can then write,
(2 32)
FIGURE 2 16 Potential energy of a
dipole in a uniform external field |
1 | 1819-1822 | ) We can then write,
(2 32)
FIGURE 2 16 Potential energy of a
dipole in a uniform external field Rationalised 2023-24
Electrostatic Potential
and Capacitance
61
EXAMPLE 2 |
1 | 1820-1823 | 32)
FIGURE 2 16 Potential energy of a
dipole in a uniform external field Rationalised 2023-24
Electrostatic Potential
and Capacitance
61
EXAMPLE 2 6
This expression can alternately be understood also from Eq |
1 | 1821-1824 | 16 Potential energy of a
dipole in a uniform external field Rationalised 2023-24
Electrostatic Potential
and Capacitance
61
EXAMPLE 2 6
This expression can alternately be understood also from Eq (2 |
1 | 1822-1825 | Rationalised 2023-24
Electrostatic Potential
and Capacitance
61
EXAMPLE 2 6
This expression can alternately be understood also from Eq (2 29) |
1 | 1823-1826 | 6
This expression can alternately be understood also from Eq (2 29) We apply Eq |
1 | 1824-1827 | (2 29) We apply Eq (2 |
1 | 1825-1828 | 29) We apply Eq (2 29) to the present system of two charges +q and –q |
1 | 1826-1829 | We apply Eq (2 29) to the present system of two charges +q and –q The
potential energy expression then reads
( )
( )
(
)
2
1
2
[
]
4
2
q
U
q V
V
a
θ
ε0
=
−
−
′
π
×
r
r
(2 |
1 | 1827-1830 | (2 29) to the present system of two charges +q and –q The
potential energy expression then reads
( )
( )
(
)
2
1
2
[
]
4
2
q
U
q V
V
a
θ
ε0
=
−
−
′
π
×
r
r
(2 33)
Here, r1 and r2 denote the position vectors of +q and –q |
1 | 1828-1831 | 29) to the present system of two charges +q and –q The
potential energy expression then reads
( )
( )
(
)
2
1
2
[
]
4
2
q
U
q V
V
a
θ
ε0
=
−
−
′
π
×
r
r
(2 33)
Here, r1 and r2 denote the position vectors of +q and –q Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1 |
1 | 1829-1832 | The
potential energy expression then reads
( )
( )
(
)
2
1
2
[
]
4
2
q
U
q V
V
a
θ
ε0
=
−
−
′
π
×
r
r
(2 33)
Here, r1 and r2 denote the position vectors of +q and –q Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1 The
displacement parallel to the force is 2a cosq |
1 | 1830-1833 | 33)
Here, r1 and r2 denote the position vectors of +q and –q Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1 The
displacement parallel to the force is 2a cosq Thus, [V(r1)–V (r2)] =
–E × 2a cosq |
1 | 1831-1834 | Now, the
potential difference between positions r1 and r2 equals the work done
in bringing a unit positive charge against field from r2 to r1 The
displacement parallel to the force is 2a cosq Thus, [V(r1)–V (r2)] =
–E × 2a cosq We thus obtain,
( )
2
2
cos
4
2
4
2
θ
θ
ε
ε
0
0
= −
−
= −
−
′
π
×
π
×
p |
1 | 1832-1835 | The
displacement parallel to the force is 2a cosq Thus, [V(r1)–V (r2)] =
–E × 2a cosq We thus obtain,
( )
2
2
cos
4
2
4
2
θ
θ
ε
ε
0
0
= −
−
= −
−
′
π
×
π
×
p E
q
q
U
pE
a
a
(2 |
1 | 1833-1836 | Thus, [V(r1)–V (r2)] =
–E × 2a cosq We thus obtain,
( )
2
2
cos
4
2
4
2
θ
θ
ε
ε
0
0
= −
−
= −
−
′
π
×
π
×
p E
q
q
U
pE
a
a
(2 34)
We note that U¢ (q) differs from U(q ) by a quantity which is just a constant
for a given dipole |
1 | 1834-1837 | We thus obtain,
( )
2
2
cos
4
2
4
2
θ
θ
ε
ε
0
0
= −
−
= −
−
′
π
×
π
×
p E
q
q
U
pE
a
a
(2 34)
We note that U¢ (q) differs from U(q ) by a quantity which is just a constant
for a given dipole Since a constant is insignificant for potential energy, we
can drop the second term in Eq |
1 | 1835-1838 | E
q
q
U
pE
a
a
(2 34)
We note that U¢ (q) differs from U(q ) by a quantity which is just a constant
for a given dipole Since a constant is insignificant for potential energy, we
can drop the second term in Eq (2 |
1 | 1836-1839 | 34)
We note that U¢ (q) differs from U(q ) by a quantity which is just a constant
for a given dipole Since a constant is insignificant for potential energy, we
can drop the second term in Eq (2 34) and it then reduces to Eq |
1 | 1837-1840 | Since a constant is insignificant for potential energy, we
can drop the second term in Eq (2 34) and it then reduces to Eq (2 |
1 | 1838-1841 | (2 34) and it then reduces to Eq (2 32) |
1 | 1839-1842 | 34) and it then reduces to Eq (2 32) We can now understand why we took q0=p/2 |
1 | 1840-1843 | (2 32) We can now understand why we took q0=p/2 In this case, the work
done against the external field E in bringing +q and – q are equal and
opposite and cancel out, i |
1 | 1841-1844 | 32) We can now understand why we took q0=p/2 In this case, the work
done against the external field E in bringing +q and – q are equal and
opposite and cancel out, i e |
1 | 1842-1845 | We can now understand why we took q0=p/2 In this case, the work
done against the external field E in bringing +q and – q are equal and
opposite and cancel out, i e , q [V (r1) – V (r2)]=0 |
1 | 1843-1846 | In this case, the work
done against the external field E in bringing +q and – q are equal and
opposite and cancel out, i e , q [V (r1) – V (r2)]=0 Example 2 |
1 | 1844-1847 | e , q [V (r1) – V (r2)]=0 Example 2 6 A molecule of a substance has a permanent electric
dipole moment of magnitude 10–29 C m |
1 | 1845-1848 | , q [V (r1) – V (r2)]=0 Example 2 6 A molecule of a substance has a permanent electric
dipole moment of magnitude 10–29 C m A mole of this substance is
polarised (at low temperature) by applying a strong electrostatic field
of magnitude 106 V m–1 |
1 | 1846-1849 | Example 2 6 A molecule of a substance has a permanent electric
dipole moment of magnitude 10–29 C m A mole of this substance is
polarised (at low temperature) by applying a strong electrostatic field
of magnitude 106 V m–1 The direction of the field is suddenly changed
by an angle of 60º |
1 | 1847-1850 | 6 A molecule of a substance has a permanent electric
dipole moment of magnitude 10–29 C m A mole of this substance is
polarised (at low temperature) by applying a strong electrostatic field
of magnitude 106 V m–1 The direction of the field is suddenly changed
by an angle of 60º Estimate the heat released by the substance in
aligning its dipoles along the new direction of the field |
1 | 1848-1851 | A mole of this substance is
polarised (at low temperature) by applying a strong electrostatic field
of magnitude 106 V m–1 The direction of the field is suddenly changed
by an angle of 60º Estimate the heat released by the substance in
aligning its dipoles along the new direction of the field For simplicity,
assume 100% polarisation of the sample |
1 | 1849-1852 | The direction of the field is suddenly changed
by an angle of 60º Estimate the heat released by the substance in
aligning its dipoles along the new direction of the field For simplicity,
assume 100% polarisation of the sample Solution Here, dipole moment of each molecules = 10–29 C m
As 1 mole of the substance contains 6 × 1023 molecules,
total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m
= 6 × 10–6
C m
Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J
Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J
Change in potential energy = –3 J – (–6J) = 3 J
So, there is loss in potential energy |
1 | 1850-1853 | Estimate the heat released by the substance in
aligning its dipoles along the new direction of the field For simplicity,
assume 100% polarisation of the sample Solution Here, dipole moment of each molecules = 10–29 C m
As 1 mole of the substance contains 6 × 1023 molecules,
total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m
= 6 × 10–6
C m
Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J
Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J
Change in potential energy = –3 J – (–6J) = 3 J
So, there is loss in potential energy This must be the energy released
by the substance in the form of heat in aligning its dipoles |
1 | 1851-1854 | For simplicity,
assume 100% polarisation of the sample Solution Here, dipole moment of each molecules = 10–29 C m
As 1 mole of the substance contains 6 × 1023 molecules,
total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m
= 6 × 10–6
C m
Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J
Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J
Change in potential energy = –3 J – (–6J) = 3 J
So, there is loss in potential energy This must be the energy released
by the substance in the form of heat in aligning its dipoles 2 |
1 | 1852-1855 | Solution Here, dipole moment of each molecules = 10–29 C m
As 1 mole of the substance contains 6 × 1023 molecules,
total dipole moment of all the molecules, p = 6 × 1023 × 10–29 C m
= 6 × 10–6
C m
Initial potential energy, Ui = –pE cos q = –6×10–6×106 cos 0° = –6 J
Final potential energy (when q = 60°), Uf = –6 × 10–6 × 106 cos 60° = –3 J
Change in potential energy = –3 J – (–6J) = 3 J
So, there is loss in potential energy This must be the energy released
by the substance in the form of heat in aligning its dipoles 2 9 ELECTROSTATICS OF CONDUCTORS
Conductors and insulators were described briefly in Chapter 1 |
1 | 1853-1856 | This must be the energy released
by the substance in the form of heat in aligning its dipoles 2 9 ELECTROSTATICS OF CONDUCTORS
Conductors and insulators were described briefly in Chapter 1 Conductors contain mobile charge carriers |
1 | 1854-1857 | 2 9 ELECTROSTATICS OF CONDUCTORS
Conductors and insulators were described briefly in Chapter 1 Conductors contain mobile charge carriers In metallic conductors, these
charge carriers are electrons |
1 | 1855-1858 | 9 ELECTROSTATICS OF CONDUCTORS
Conductors and insulators were described briefly in Chapter 1 Conductors contain mobile charge carriers In metallic conductors, these
charge carriers are electrons In a metal, the outer (valence) electrons
part away from their atoms and are free to move |
1 | 1856-1859 | Conductors contain mobile charge carriers In metallic conductors, these
charge carriers are electrons In a metal, the outer (valence) electrons
part away from their atoms and are free to move These electrons are free
within the metal but not free to leave the metal |
1 | 1857-1860 | In metallic conductors, these
charge carriers are electrons In a metal, the outer (valence) electrons
part away from their atoms and are free to move These electrons are free
within the metal but not free to leave the metal The free electrons form a
kind of ‘gas’; they collide with each other and with the ions, and move
randomly in different directions |
1 | 1858-1861 | In a metal, the outer (valence) electrons
part away from their atoms and are free to move These electrons are free
within the metal but not free to leave the metal The free electrons form a
kind of ‘gas’; they collide with each other and with the ions, and move
randomly in different directions In an external electric field, they drift
against the direction of the field |
1 | 1859-1862 | These electrons are free
within the metal but not free to leave the metal The free electrons form a
kind of ‘gas’; they collide with each other and with the ions, and move
randomly in different directions In an external electric field, they drift
against the direction of the field The positive ions made up of the nuclei
and the bound electrons remain held in their fixed positions |
1 | 1860-1863 | The free electrons form a
kind of ‘gas’; they collide with each other and with the ions, and move
randomly in different directions In an external electric field, they drift
against the direction of the field The positive ions made up of the nuclei
and the bound electrons remain held in their fixed positions In electrolytic
conductors, the charge carriers are both positive and negative ions; but
Rationalised 2023-24
Physics
62
the situation in this case is more involved – the movement of the charge
carriers is affected both by the external electric field as also by the
so-called chemical forces (see Chapter 3) |
1 | 1861-1864 | In an external electric field, they drift
against the direction of the field The positive ions made up of the nuclei
and the bound electrons remain held in their fixed positions In electrolytic
conductors, the charge carriers are both positive and negative ions; but
Rationalised 2023-24
Physics
62
the situation in this case is more involved – the movement of the charge
carriers is affected both by the external electric field as also by the
so-called chemical forces (see Chapter 3) We shall restrict our discussion
to metallic solid conductors |
1 | 1862-1865 | The positive ions made up of the nuclei
and the bound electrons remain held in their fixed positions In electrolytic
conductors, the charge carriers are both positive and negative ions; but
Rationalised 2023-24
Physics
62
the situation in this case is more involved – the movement of the charge
carriers is affected both by the external electric field as also by the
so-called chemical forces (see Chapter 3) We shall restrict our discussion
to metallic solid conductors Let us note important results regarding
electrostatics of conductors |
1 | 1863-1866 | In electrolytic
conductors, the charge carriers are both positive and negative ions; but
Rationalised 2023-24
Physics
62
the situation in this case is more involved – the movement of the charge
carriers is affected both by the external electric field as also by the
so-called chemical forces (see Chapter 3) We shall restrict our discussion
to metallic solid conductors Let us note important results regarding
electrostatics of conductors 1 |
1 | 1864-1867 | We shall restrict our discussion
to metallic solid conductors Let us note important results regarding
electrostatics of conductors 1 Inside a conductor, electrostatic field is zero
Consider a conductor, neutral or charged |
1 | 1865-1868 | Let us note important results regarding
electrostatics of conductors 1 Inside a conductor, electrostatic field is zero
Consider a conductor, neutral or charged There may also be an external
electrostatic field |
1 | 1866-1869 | 1 Inside a conductor, electrostatic field is zero
Consider a conductor, neutral or charged There may also be an external
electrostatic field In the static situation, when there is no current inside
or on the surface of the conductor, the electric field is zero everywhere
inside the conductor |
1 | 1867-1870 | Inside a conductor, electrostatic field is zero
Consider a conductor, neutral or charged There may also be an external
electrostatic field In the static situation, when there is no current inside
or on the surface of the conductor, the electric field is zero everywhere
inside the conductor This fact can be taken as the defining property of a
conductor |
1 | 1868-1871 | There may also be an external
electrostatic field In the static situation, when there is no current inside
or on the surface of the conductor, the electric field is zero everywhere
inside the conductor This fact can be taken as the defining property of a
conductor A conductor has free electrons |
1 | 1869-1872 | In the static situation, when there is no current inside
or on the surface of the conductor, the electric field is zero everywhere
inside the conductor This fact can be taken as the defining property of a
conductor A conductor has free electrons As long as electric field is not
zero, the free charge carriers would experience force and drift |
1 | 1870-1873 | This fact can be taken as the defining property of a
conductor A conductor has free electrons As long as electric field is not
zero, the free charge carriers would experience force and drift In the
static situation, the free charges have so distributed themselves that the
electric field is zero everywhere inside |
1 | 1871-1874 | A conductor has free electrons As long as electric field is not
zero, the free charge carriers would experience force and drift In the
static situation, the free charges have so distributed themselves that the
electric field is zero everywhere inside Electrostatic field is zero inside a
conductor |
1 | 1872-1875 | As long as electric field is not
zero, the free charge carriers would experience force and drift In the
static situation, the free charges have so distributed themselves that the
electric field is zero everywhere inside Electrostatic field is zero inside a
conductor 2 |
1 | 1873-1876 | In the
static situation, the free charges have so distributed themselves that the
electric field is zero everywhere inside Electrostatic field is zero inside a
conductor 2 At the surface of a charged conductor, electrostatic field
must be normal to the surface at every point
If E were not normal to the surface, it would have some non-zero
component along the surface |
1 | 1874-1877 | Electrostatic field is zero inside a
conductor 2 At the surface of a charged conductor, electrostatic field
must be normal to the surface at every point
If E were not normal to the surface, it would have some non-zero
component along the surface Free charges on the surface of the conductor
would then experience force and move |
1 | 1875-1878 | 2 At the surface of a charged conductor, electrostatic field
must be normal to the surface at every point
If E were not normal to the surface, it would have some non-zero
component along the surface Free charges on the surface of the conductor
would then experience force and move In the static situation, therefore,
E should have no tangential component |
1 | 1876-1879 | At the surface of a charged conductor, electrostatic field
must be normal to the surface at every point
If E were not normal to the surface, it would have some non-zero
component along the surface Free charges on the surface of the conductor
would then experience force and move In the static situation, therefore,
E should have no tangential component Thus electrostatic field at the
surface of a charged conductor must be normal to the surface at every
point |
1 | 1877-1880 | Free charges on the surface of the conductor
would then experience force and move In the static situation, therefore,
E should have no tangential component Thus electrostatic field at the
surface of a charged conductor must be normal to the surface at every
point (For a conductor without any surface charge density, field is zero
even at the surface |
1 | 1878-1881 | In the static situation, therefore,
E should have no tangential component Thus electrostatic field at the
surface of a charged conductor must be normal to the surface at every
point (For a conductor without any surface charge density, field is zero
even at the surface ) See result 5 |
1 | 1879-1882 | Thus electrostatic field at the
surface of a charged conductor must be normal to the surface at every
point (For a conductor without any surface charge density, field is zero
even at the surface ) See result 5 3 |
1 | 1880-1883 | (For a conductor without any surface charge density, field is zero
even at the surface ) See result 5 3 The interior of a conductor can have no excess charge in
the static situation
A neutral conductor has equal amounts of positive and negative charges
in every small volume or surface element |
1 | 1881-1884 | ) See result 5 3 The interior of a conductor can have no excess charge in
the static situation
A neutral conductor has equal amounts of positive and negative charges
in every small volume or surface element When the conductor is charged,
the excess charge can reside only on the surface in the static situation |
1 | 1882-1885 | 3 The interior of a conductor can have no excess charge in
the static situation
A neutral conductor has equal amounts of positive and negative charges
in every small volume or surface element When the conductor is charged,
the excess charge can reside only on the surface in the static situation This follows from the Gauss’s law |
1 | 1883-1886 | The interior of a conductor can have no excess charge in
the static situation
A neutral conductor has equal amounts of positive and negative charges
in every small volume or surface element When the conductor is charged,
the excess charge can reside only on the surface in the static situation This follows from the Gauss’s law Consider any arbitrary volume element
v inside a conductor |
1 | 1884-1887 | When the conductor is charged,
the excess charge can reside only on the surface in the static situation This follows from the Gauss’s law Consider any arbitrary volume element
v inside a conductor On the closed surface S bounding the volume
element v, electrostatic field is zero |
1 | 1885-1888 | This follows from the Gauss’s law Consider any arbitrary volume element
v inside a conductor On the closed surface S bounding the volume
element v, electrostatic field is zero Thus the total electric flux through S
is zero |
1 | 1886-1889 | Consider any arbitrary volume element
v inside a conductor On the closed surface S bounding the volume
element v, electrostatic field is zero Thus the total electric flux through S
is zero Hence, by Gauss’s law, there is no net charge enclosed by S |
1 | 1887-1890 | On the closed surface S bounding the volume
element v, electrostatic field is zero Thus the total electric flux through S
is zero Hence, by Gauss’s law, there is no net charge enclosed by S But
the surface S can be made as small as you like, i |
1 | 1888-1891 | Thus the total electric flux through S
is zero Hence, by Gauss’s law, there is no net charge enclosed by S But
the surface S can be made as small as you like, i e |
1 | 1889-1892 | Hence, by Gauss’s law, there is no net charge enclosed by S But
the surface S can be made as small as you like, i e , the volume v can be
made vanishingly small |
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