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1 | 790-793 | In this example, the ratio OP/OB is quite large (= 60) Thus, we can
expect to get approximately the same result as above by directly using
the formula for electric field at a far-away point on the axis of a dipole For a dipole consisting of charges ± q, 2a distance apart, the electric
field at a distance r from the centre on the axis of the dipole has a
magnitude
E
p
r
=
2
4
0
3
πε
(r/a >> 1)
where p = 2a q is the magnitude of the dipole moment The direction of electric field on the dipole axis is always along the
direction of the dipole moment vector (i |
1 | 791-794 | Thus, we can
expect to get approximately the same result as above by directly using
the formula for electric field at a far-away point on the axis of a dipole For a dipole consisting of charges ± q, 2a distance apart, the electric
field at a distance r from the centre on the axis of the dipole has a
magnitude
E
p
r
=
2
4
0
3
πε
(r/a >> 1)
where p = 2a q is the magnitude of the dipole moment The direction of electric field on the dipole axis is always along the
direction of the dipole moment vector (i e |
1 | 792-795 | For a dipole consisting of charges ± q, 2a distance apart, the electric
field at a distance r from the centre on the axis of the dipole has a
magnitude
E
p
r
=
2
4
0
3
πε
(r/a >> 1)
where p = 2a q is the magnitude of the dipole moment The direction of electric field on the dipole axis is always along the
direction of the dipole moment vector (i e , from –q to q) |
1 | 793-796 | The direction of electric field on the dipole axis is always along the
direction of the dipole moment vector (i e , from –q to q) Here,
p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m
Therefore,
E =
8
12
2
1
2
2
5 10
C m
4 (8 |
1 | 794-797 | e , from –q to q) Here,
p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m
Therefore,
E =
8
12
2
1
2
2
5 10
C m
4 (8 854 10
C N
m
)
−
−
−
−
×
×
π
×
3
6
3
1
(15)
10
m
−
×
×
= 2 |
1 | 795-798 | , from –q to q) Here,
p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m
Therefore,
E =
8
12
2
1
2
2
5 10
C m
4 (8 854 10
C N
m
)
−
−
−
−
×
×
π
×
3
6
3
1
(15)
10
m
−
×
×
= 2 6 × 105 N C–1
along the dipole moment direction AB, which is close to the result
obtained earlier |
1 | 796-799 | Here,
p =10–5 C × 5 × 10–3 m = 5 × 10–8 C m
Therefore,
E =
8
12
2
1
2
2
5 10
C m
4 (8 854 10
C N
m
)
−
−
−
−
×
×
π
×
3
6
3
1
(15)
10
m
−
×
×
= 2 6 × 105 N C–1
along the dipole moment direction AB, which is close to the result
obtained earlier (b) Field at Q due to charge + 10 mC at B
=
5
12
2
1
2
10
C
4
(8 |
1 | 797-800 | 854 10
C N
m
)
−
−
−
−
×
×
π
×
3
6
3
1
(15)
10
m
−
×
×
= 2 6 × 105 N C–1
along the dipole moment direction AB, which is close to the result
obtained earlier (b) Field at Q due to charge + 10 mC at B
=
5
12
2
1
2
10
C
4
(8 854
10
C
N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 |
1 | 798-801 | 6 × 105 N C–1
along the dipole moment direction AB, which is close to the result
obtained earlier (b) Field at Q due to charge + 10 mC at B
=
5
12
2
1
2
10
C
4
(8 854
10
C
N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ]
10
m
−
+
×
×
= 3 |
1 | 799-802 | (b) Field at Q due to charge + 10 mC at B
=
5
12
2
1
2
10
C
4
(8 854
10
C
N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ]
10
m
−
+
×
×
= 3 99 × 106 N C–1 along BQ
Field at Q due to charge –10 mC at A
=
5
12
2
1
2
10
C
4
(8 |
1 | 800-803 | 854
10
C
N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ]
10
m
−
+
×
×
= 3 99 × 106 N C–1 along BQ
Field at Q due to charge –10 mC at A
=
5
12
2
1
2
10
C
4
(8 854 10
C N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 |
1 | 801-804 | 25) ]
10
m
−
+
×
×
= 3 99 × 106 N C–1 along BQ
Field at Q due to charge –10 mC at A
=
5
12
2
1
2
10
C
4
(8 854 10
C N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ] 10
m
−
+
×
×
= 3 |
1 | 802-805 | 99 × 106 N C–1 along BQ
Field at Q due to charge –10 mC at A
=
5
12
2
1
2
10
C
4
(8 854 10
C N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ] 10
m
−
+
×
×
= 3 99 × 106 N C–1 along QA |
1 | 803-806 | 854 10
C N
m
)
−
−
−
−
π
×
2
2
4
2
1
[15
(0 25) ] 10
m
−
+
×
×
= 3 99 × 106 N C–1 along QA Clearly, the components of these two forces with equal magnitudes
cancel along the direction OQ but add up along the direction parallel
to BA |
1 | 804-807 | 25) ] 10
m
−
+
×
×
= 3 99 × 106 N C–1 along QA Clearly, the components of these two forces with equal magnitudes
cancel along the direction OQ but add up along the direction parallel
to BA Therefore, the resultant electric field at Q due to the two
charges at A and B is
= 2 ×
6
–1
2
2
0 |
1 | 805-808 | 99 × 106 N C–1 along QA Clearly, the components of these two forces with equal magnitudes
cancel along the direction OQ but add up along the direction parallel
to BA Therefore, the resultant electric field at Q due to the two
charges at A and B is
= 2 ×
6
–1
2
2
0 25
3 |
1 | 806-809 | Clearly, the components of these two forces with equal magnitudes
cancel along the direction OQ but add up along the direction parallel
to BA Therefore, the resultant electric field at Q due to the two
charges at A and B is
= 2 ×
6
–1
2
2
0 25
3 99
10 N C
15
(0 |
1 | 807-810 | Therefore, the resultant electric field at Q due to the two
charges at A and B is
= 2 ×
6
–1
2
2
0 25
3 99
10 N C
15
(0 25)
×
×
+
along BA
= 1 |
1 | 808-811 | 25
3 99
10 N C
15
(0 25)
×
×
+
along BA
= 1 33 × 105 N C–1 along BA |
1 | 809-812 | 99
10 N C
15
(0 25)
×
×
+
along BA
= 1 33 × 105 N C–1 along BA As in (a), we can expect to get approximately the same result by
directly using the formula for dipole field at a point on the normal to
the axis of the dipole:
Rationalised 2023-24
Electric Charges
and Fields
27
EXAMPLE 1 |
1 | 810-813 | 25)
×
×
+
along BA
= 1 33 × 105 N C–1 along BA As in (a), we can expect to get approximately the same result by
directly using the formula for dipole field at a point on the normal to
the axis of the dipole:
Rationalised 2023-24
Electric Charges
and Fields
27
EXAMPLE 1 9
E
p
r
= 4
3
π
0ε
(r/a >> 1)
8
12
2
–1
–2
5 10
Cm
4
(8 |
1 | 811-814 | 33 × 105 N C–1 along BA As in (a), we can expect to get approximately the same result by
directly using the formula for dipole field at a point on the normal to
the axis of the dipole:
Rationalised 2023-24
Electric Charges
and Fields
27
EXAMPLE 1 9
E
p
r
= 4
3
π
0ε
(r/a >> 1)
8
12
2
–1
–2
5 10
Cm
4
(8 854 10
C N
m
)
−
−
×
=
π
×
3
6
3
1
(15)
10
m
−
×
×
= 1 |
1 | 812-815 | As in (a), we can expect to get approximately the same result by
directly using the formula for dipole field at a point on the normal to
the axis of the dipole:
Rationalised 2023-24
Electric Charges
and Fields
27
EXAMPLE 1 9
E
p
r
= 4
3
π
0ε
(r/a >> 1)
8
12
2
–1
–2
5 10
Cm
4
(8 854 10
C N
m
)
−
−
×
=
π
×
3
6
3
1
(15)
10
m
−
×
×
= 1 33 × 105 N C–1 |
1 | 813-816 | 9
E
p
r
= 4
3
π
0ε
(r/a >> 1)
8
12
2
–1
–2
5 10
Cm
4
(8 854 10
C N
m
)
−
−
×
=
π
×
3
6
3
1
(15)
10
m
−
×
×
= 1 33 × 105 N C–1 The direction of electric field in this case is opposite to the direction
of the dipole moment vector |
1 | 814-817 | 854 10
C N
m
)
−
−
×
=
π
×
3
6
3
1
(15)
10
m
−
×
×
= 1 33 × 105 N C–1 The direction of electric field in this case is opposite to the direction
of the dipole moment vector Again, the result agrees with that obtained
before |
1 | 815-818 | 33 × 105 N C–1 The direction of electric field in this case is opposite to the direction
of the dipole moment vector Again, the result agrees with that obtained
before 1 |
1 | 816-819 | The direction of electric field in this case is opposite to the direction
of the dipole moment vector Again, the result agrees with that obtained
before 1 11 DIPOLE IN A UNIFORM EXTERNAL FIELD
Consider a permanent dipole of dipole moment p in a uniform
external field E, as shown in Fig |
1 | 817-820 | Again, the result agrees with that obtained
before 1 11 DIPOLE IN A UNIFORM EXTERNAL FIELD
Consider a permanent dipole of dipole moment p in a uniform
external field E, as shown in Fig 1 |
1 | 818-821 | 1 11 DIPOLE IN A UNIFORM EXTERNAL FIELD
Consider a permanent dipole of dipole moment p in a uniform
external field E, as shown in Fig 1 19 |
1 | 819-822 | 11 DIPOLE IN A UNIFORM EXTERNAL FIELD
Consider a permanent dipole of dipole moment p in a uniform
external field E, as shown in Fig 1 19 (By permanent dipole, we
mean that p exists irrespective of E; it has not been induced by E |
1 | 820-823 | 1 19 (By permanent dipole, we
mean that p exists irrespective of E; it has not been induced by E )
There is a force qE on q and a force –qE on –q |
1 | 821-824 | 19 (By permanent dipole, we
mean that p exists irrespective of E; it has not been induced by E )
There is a force qE on q and a force –qE on –q The net force on
the dipole is zero, since E is uniform |
1 | 822-825 | (By permanent dipole, we
mean that p exists irrespective of E; it has not been induced by E )
There is a force qE on q and a force –qE on –q The net force on
the dipole is zero, since E is uniform However, the charges are
separated, so the forces act at different points, resulting in a torque
on the dipole |
1 | 823-826 | )
There is a force qE on q and a force –qE on –q The net force on
the dipole is zero, since E is uniform However, the charges are
separated, so the forces act at different points, resulting in a torque
on the dipole When the net force is zero, the torque (couple) is
independent of the origin |
1 | 824-827 | The net force on
the dipole is zero, since E is uniform However, the charges are
separated, so the forces act at different points, resulting in a torque
on the dipole When the net force is zero, the torque (couple) is
independent of the origin Its magnitude equals the magnitude of
each force multiplied by the arm of the couple (perpendicular
distance between the two antiparallel forces) |
1 | 825-828 | However, the charges are
separated, so the forces act at different points, resulting in a torque
on the dipole When the net force is zero, the torque (couple) is
independent of the origin Its magnitude equals the magnitude of
each force multiplied by the arm of the couple (perpendicular
distance between the two antiparallel forces) Magnitude of torque = q E × 2 a sinq
= 2 q a E sinq
Its direction is normal to the plane of the paper, coming out of it |
1 | 826-829 | When the net force is zero, the torque (couple) is
independent of the origin Its magnitude equals the magnitude of
each force multiplied by the arm of the couple (perpendicular
distance between the two antiparallel forces) Magnitude of torque = q E × 2 a sinq
= 2 q a E sinq
Its direction is normal to the plane of the paper, coming out of it The magnitude of p × E is also p E sinq and its direction
is normal to the paper, coming out of it |
1 | 827-830 | Its magnitude equals the magnitude of
each force multiplied by the arm of the couple (perpendicular
distance between the two antiparallel forces) Magnitude of torque = q E × 2 a sinq
= 2 q a E sinq
Its direction is normal to the plane of the paper, coming out of it The magnitude of p × E is also p E sinq and its direction
is normal to the paper, coming out of it Thus,
ttttt = p × E
(1 |
1 | 828-831 | Magnitude of torque = q E × 2 a sinq
= 2 q a E sinq
Its direction is normal to the plane of the paper, coming out of it The magnitude of p × E is also p E sinq and its direction
is normal to the paper, coming out of it Thus,
ttttt = p × E
(1 22)
This torque will tend to align the dipole with the field
E |
1 | 829-832 | The magnitude of p × E is also p E sinq and its direction
is normal to the paper, coming out of it Thus,
ttttt = p × E
(1 22)
This torque will tend to align the dipole with the field
E When p is aligned with E, the torque is zero |
1 | 830-833 | Thus,
ttttt = p × E
(1 22)
This torque will tend to align the dipole with the field
E When p is aligned with E, the torque is zero What happens if the field is not uniform |
1 | 831-834 | 22)
This torque will tend to align the dipole with the field
E When p is aligned with E, the torque is zero What happens if the field is not uniform In that case,
the net force will evidently be non-zero |
1 | 832-835 | When p is aligned with E, the torque is zero What happens if the field is not uniform In that case,
the net force will evidently be non-zero In addition there
will, in general, be a torque on the system as before |
1 | 833-836 | What happens if the field is not uniform In that case,
the net force will evidently be non-zero In addition there
will, in general, be a torque on the system as before The
general case is involved, so let us consider the simpler
situations when p is parallel to E or antiparallel to E |
1 | 834-837 | In that case,
the net force will evidently be non-zero In addition there
will, in general, be a torque on the system as before The
general case is involved, so let us consider the simpler
situations when p is parallel to E or antiparallel to E In
either case, the net torque is zero, but there is a net force
on the dipole if E is not uniform |
1 | 835-838 | In addition there
will, in general, be a torque on the system as before The
general case is involved, so let us consider the simpler
situations when p is parallel to E or antiparallel to E In
either case, the net torque is zero, but there is a net force
on the dipole if E is not uniform Figure 1 |
1 | 836-839 | The
general case is involved, so let us consider the simpler
situations when p is parallel to E or antiparallel to E In
either case, the net torque is zero, but there is a net force
on the dipole if E is not uniform Figure 1 20 is self-explanatory |
1 | 837-840 | In
either case, the net torque is zero, but there is a net force
on the dipole if E is not uniform Figure 1 20 is self-explanatory It is easily seen that
when p is parallel to E, the dipole has a net force in the
direction of increasing field |
1 | 838-841 | Figure 1 20 is self-explanatory It is easily seen that
when p is parallel to E, the dipole has a net force in the
direction of increasing field When p is antiparallel to E,
the net force on the dipole is in the direction of decreasing
field |
1 | 839-842 | 20 is self-explanatory It is easily seen that
when p is parallel to E, the dipole has a net force in the
direction of increasing field When p is antiparallel to E,
the net force on the dipole is in the direction of decreasing
field In general, the force depends on the orientation of p
with respect to E |
1 | 840-843 | It is easily seen that
when p is parallel to E, the dipole has a net force in the
direction of increasing field When p is antiparallel to E,
the net force on the dipole is in the direction of decreasing
field In general, the force depends on the orientation of p
with respect to E This brings us to a common observation in frictional
electricity |
1 | 841-844 | When p is antiparallel to E,
the net force on the dipole is in the direction of decreasing
field In general, the force depends on the orientation of p
with respect to E This brings us to a common observation in frictional
electricity A comb run through dry hair attracts pieces of
paper |
1 | 842-845 | In general, the force depends on the orientation of p
with respect to E This brings us to a common observation in frictional
electricity A comb run through dry hair attracts pieces of
paper The comb, as we know, acquires charge through
friction |
1 | 843-846 | This brings us to a common observation in frictional
electricity A comb run through dry hair attracts pieces of
paper The comb, as we know, acquires charge through
friction But the paper is not charged |
1 | 844-847 | A comb run through dry hair attracts pieces of
paper The comb, as we know, acquires charge through
friction But the paper is not charged What then explains
the attractive force |
1 | 845-848 | The comb, as we know, acquires charge through
friction But the paper is not charged What then explains
the attractive force Taking the clue from the preceding
FIGURE 1 |
1 | 846-849 | But the paper is not charged What then explains
the attractive force Taking the clue from the preceding
FIGURE 1 19 Dipole in a
uniform electric field |
1 | 847-850 | What then explains
the attractive force Taking the clue from the preceding
FIGURE 1 19 Dipole in a
uniform electric field FIGURE 1 |
1 | 848-851 | Taking the clue from the preceding
FIGURE 1 19 Dipole in a
uniform electric field FIGURE 1 20 Electric force on a
dipole: (a) E parallel to p, (b) E
antiparallel to p |
1 | 849-852 | 19 Dipole in a
uniform electric field FIGURE 1 20 Electric force on a
dipole: (a) E parallel to p, (b) E
antiparallel to p Rationalised 2023-24
28
Physics
discussion, the charged comb ‘polarises’ the piece of paper, i |
1 | 850-853 | FIGURE 1 20 Electric force on a
dipole: (a) E parallel to p, (b) E
antiparallel to p Rationalised 2023-24
28
Physics
discussion, the charged comb ‘polarises’ the piece of paper, i e |
1 | 851-854 | 20 Electric force on a
dipole: (a) E parallel to p, (b) E
antiparallel to p Rationalised 2023-24
28
Physics
discussion, the charged comb ‘polarises’ the piece of paper, i e , induces
a net dipole moment in the direction of field |
1 | 852-855 | Rationalised 2023-24
28
Physics
discussion, the charged comb ‘polarises’ the piece of paper, i e , induces
a net dipole moment in the direction of field Further, the electric field
due to the comb is not uniform |
1 | 853-856 | e , induces
a net dipole moment in the direction of field Further, the electric field
due to the comb is not uniform This non-uniformity of the field makes a
dipole to experience a net force on it |
1 | 854-857 | , induces
a net dipole moment in the direction of field Further, the electric field
due to the comb is not uniform This non-uniformity of the field makes a
dipole to experience a net force on it In this situation, it is easily seen
that the paper should move in the direction of the comb |
1 | 855-858 | Further, the electric field
due to the comb is not uniform This non-uniformity of the field makes a
dipole to experience a net force on it In this situation, it is easily seen
that the paper should move in the direction of the comb 1 |
1 | 856-859 | This non-uniformity of the field makes a
dipole to experience a net force on it In this situation, it is easily seen
that the paper should move in the direction of the comb 1 12 CONTINUOUS CHARGE DISTRIBUTION
We have so far dealt with charge configurations involving discrete charges
q1, q2, |
1 | 857-860 | In this situation, it is easily seen
that the paper should move in the direction of the comb 1 12 CONTINUOUS CHARGE DISTRIBUTION
We have so far dealt with charge configurations involving discrete charges
q1, q2, , qn |
1 | 858-861 | 1 12 CONTINUOUS CHARGE DISTRIBUTION
We have so far dealt with charge configurations involving discrete charges
q1, q2, , qn One reason why we restricted to discrete charges is that the
mathematical treatment is simpler and does not involve calculus |
1 | 859-862 | 12 CONTINUOUS CHARGE DISTRIBUTION
We have so far dealt with charge configurations involving discrete charges
q1, q2, , qn One reason why we restricted to discrete charges is that the
mathematical treatment is simpler and does not involve calculus For
many purposes, however, it is impractical to work in terms of discrete
charges and we need to work with continuous charge distributions |
1 | 860-863 | , qn One reason why we restricted to discrete charges is that the
mathematical treatment is simpler and does not involve calculus For
many purposes, however, it is impractical to work in terms of discrete
charges and we need to work with continuous charge distributions For
example, on the surface of a charged conductor, it is impractical to specify
the charge distribution in terms of the locations of the microscopic charged
constituents |
1 | 861-864 | One reason why we restricted to discrete charges is that the
mathematical treatment is simpler and does not involve calculus For
many purposes, however, it is impractical to work in terms of discrete
charges and we need to work with continuous charge distributions For
example, on the surface of a charged conductor, it is impractical to specify
the charge distribution in terms of the locations of the microscopic charged
constituents It is more feasible to consider an area element DS (Fig |
1 | 862-865 | For
many purposes, however, it is impractical to work in terms of discrete
charges and we need to work with continuous charge distributions For
example, on the surface of a charged conductor, it is impractical to specify
the charge distribution in terms of the locations of the microscopic charged
constituents It is more feasible to consider an area element DS (Fig 1 |
1 | 863-866 | For
example, on the surface of a charged conductor, it is impractical to specify
the charge distribution in terms of the locations of the microscopic charged
constituents It is more feasible to consider an area element DS (Fig 1 21)
on the surface of the conductor (which is very small on the macroscopic
scale but big enough to include a very large number of electrons) and
specify the charge DQ on that element |
1 | 864-867 | It is more feasible to consider an area element DS (Fig 1 21)
on the surface of the conductor (which is very small on the macroscopic
scale but big enough to include a very large number of electrons) and
specify the charge DQ on that element We then define a surface charge
density s at the area element by
SQ
σ
= ∆∆
(1 |
1 | 865-868 | 1 21)
on the surface of the conductor (which is very small on the macroscopic
scale but big enough to include a very large number of electrons) and
specify the charge DQ on that element We then define a surface charge
density s at the area element by
SQ
σ
= ∆∆
(1 23)
We can do this at different points on the conductor and thus arrive at
a continuous function s, called the surface charge density |
1 | 866-869 | 21)
on the surface of the conductor (which is very small on the macroscopic
scale but big enough to include a very large number of electrons) and
specify the charge DQ on that element We then define a surface charge
density s at the area element by
SQ
σ
= ∆∆
(1 23)
We can do this at different points on the conductor and thus arrive at
a continuous function s, called the surface charge density The surface
charge density s so defined ignores the quantisation of charge and the
discontinuity in charge distribution at the microscopic level* |
1 | 867-870 | We then define a surface charge
density s at the area element by
SQ
σ
= ∆∆
(1 23)
We can do this at different points on the conductor and thus arrive at
a continuous function s, called the surface charge density The surface
charge density s so defined ignores the quantisation of charge and the
discontinuity in charge distribution at the microscopic level* s represents
macroscopic surface charge density, which in a sense, is a smoothed out
average of the microscopic charge density over an area element DS which,
as said before, is large microscopically but small macroscopically |
1 | 868-871 | 23)
We can do this at different points on the conductor and thus arrive at
a continuous function s, called the surface charge density The surface
charge density s so defined ignores the quantisation of charge and the
discontinuity in charge distribution at the microscopic level* s represents
macroscopic surface charge density, which in a sense, is a smoothed out
average of the microscopic charge density over an area element DS which,
as said before, is large microscopically but small macroscopically The
units for s are C/m2 |
1 | 869-872 | The surface
charge density s so defined ignores the quantisation of charge and the
discontinuity in charge distribution at the microscopic level* s represents
macroscopic surface charge density, which in a sense, is a smoothed out
average of the microscopic charge density over an area element DS which,
as said before, is large microscopically but small macroscopically The
units for s are C/m2 Similar considerations apply for a line charge distribution and a volume
charge distribution |
1 | 870-873 | s represents
macroscopic surface charge density, which in a sense, is a smoothed out
average of the microscopic charge density over an area element DS which,
as said before, is large microscopically but small macroscopically The
units for s are C/m2 Similar considerations apply for a line charge distribution and a volume
charge distribution The linear charge density l of a wire is defined by
lQ
λ
=∆
∆
(1 |
1 | 871-874 | The
units for s are C/m2 Similar considerations apply for a line charge distribution and a volume
charge distribution The linear charge density l of a wire is defined by
lQ
λ
=∆
∆
(1 24)
where Dl is a small line element of wire on the macroscopic scale that,
however, includes a large number of microscopic charged constituents,
and DQ is the charge contained in that line element |
1 | 872-875 | Similar considerations apply for a line charge distribution and a volume
charge distribution The linear charge density l of a wire is defined by
lQ
λ
=∆
∆
(1 24)
where Dl is a small line element of wire on the macroscopic scale that,
however, includes a large number of microscopic charged constituents,
and DQ is the charge contained in that line element The units for l are
C/m |
1 | 873-876 | The linear charge density l of a wire is defined by
lQ
λ
=∆
∆
(1 24)
where Dl is a small line element of wire on the macroscopic scale that,
however, includes a large number of microscopic charged constituents,
and DQ is the charge contained in that line element The units for l are
C/m The volume charge density (sometimes simply called charge density)
is defined in a similar manner:
VQ
ρ
= ∆∆
(1 |
1 | 874-877 | 24)
where Dl is a small line element of wire on the macroscopic scale that,
however, includes a large number of microscopic charged constituents,
and DQ is the charge contained in that line element The units for l are
C/m The volume charge density (sometimes simply called charge density)
is defined in a similar manner:
VQ
ρ
= ∆∆
(1 25)
where DQ is the charge included in the macroscopically small volume
element DV that includes a large number of microscopic charged
constituents |
1 | 875-878 | The units for l are
C/m The volume charge density (sometimes simply called charge density)
is defined in a similar manner:
VQ
ρ
= ∆∆
(1 25)
where DQ is the charge included in the macroscopically small volume
element DV that includes a large number of microscopic charged
constituents The units for r are C/m3 |
1 | 876-879 | The volume charge density (sometimes simply called charge density)
is defined in a similar manner:
VQ
ρ
= ∆∆
(1 25)
where DQ is the charge included in the macroscopically small volume
element DV that includes a large number of microscopic charged
constituents The units for r are C/m3 The notion of continuous charge distribution is similar to that we
adopt for continuous mass distribution in mechanics |
1 | 877-880 | 25)
where DQ is the charge included in the macroscopically small volume
element DV that includes a large number of microscopic charged
constituents The units for r are C/m3 The notion of continuous charge distribution is similar to that we
adopt for continuous mass distribution in mechanics When we refer to
FIGURE 1 |
1 | 878-881 | The units for r are C/m3 The notion of continuous charge distribution is similar to that we
adopt for continuous mass distribution in mechanics When we refer to
FIGURE 1 21
Definition of linear,
surface and volume
charge densities |
1 | 879-882 | The notion of continuous charge distribution is similar to that we
adopt for continuous mass distribution in mechanics When we refer to
FIGURE 1 21
Definition of linear,
surface and volume
charge densities In each case, the
element (Dl, DS, DV)
chosen is small on
the macroscopic
scale but contains
a very large number
of microscopic
constituents |
1 | 880-883 | When we refer to
FIGURE 1 21
Definition of linear,
surface and volume
charge densities In each case, the
element (Dl, DS, DV)
chosen is small on
the macroscopic
scale but contains
a very large number
of microscopic
constituents *
At the microscopic level, charge distribution is discontinuous, because they are
discrete charges separated by intervening space where there is no charge |
1 | 881-884 | 21
Definition of linear,
surface and volume
charge densities In each case, the
element (Dl, DS, DV)
chosen is small on
the macroscopic
scale but contains
a very large number
of microscopic
constituents *
At the microscopic level, charge distribution is discontinuous, because they are
discrete charges separated by intervening space where there is no charge Rationalised 2023-24
Electric Charges
and Fields
29
the density of a liquid, we are referring to its macroscopic density |
1 | 882-885 | In each case, the
element (Dl, DS, DV)
chosen is small on
the macroscopic
scale but contains
a very large number
of microscopic
constituents *
At the microscopic level, charge distribution is discontinuous, because they are
discrete charges separated by intervening space where there is no charge Rationalised 2023-24
Electric Charges
and Fields
29
the density of a liquid, we are referring to its macroscopic density We
regard it as a continuous fluid and ignore its discrete molecular
constitution |
1 | 883-886 | *
At the microscopic level, charge distribution is discontinuous, because they are
discrete charges separated by intervening space where there is no charge Rationalised 2023-24
Electric Charges
and Fields
29
the density of a liquid, we are referring to its macroscopic density We
regard it as a continuous fluid and ignore its discrete molecular
constitution The field due to a continuous charge distribution can be obtained in
much the same way as for a system of discrete charges, Eq |
1 | 884-887 | Rationalised 2023-24
Electric Charges
and Fields
29
the density of a liquid, we are referring to its macroscopic density We
regard it as a continuous fluid and ignore its discrete molecular
constitution The field due to a continuous charge distribution can be obtained in
much the same way as for a system of discrete charges, Eq (1 |
1 | 885-888 | We
regard it as a continuous fluid and ignore its discrete molecular
constitution The field due to a continuous charge distribution can be obtained in
much the same way as for a system of discrete charges, Eq (1 10) |
1 | 886-889 | The field due to a continuous charge distribution can be obtained in
much the same way as for a system of discrete charges, Eq (1 10) Suppose
a continuous charge distribution in space has a charge density r |
1 | 887-890 | (1 10) Suppose
a continuous charge distribution in space has a charge density r Choose
any convenient origin O and let the position vector of any point in the
charge distribution be r |
1 | 888-891 | 10) Suppose
a continuous charge distribution in space has a charge density r 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 |
1 | 889-892 | Suppose
a continuous charge distribution in space has a charge density r 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 |
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