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1 | 490-493 | Example 1 7 An electron falls through a distance of 1 5 cm in a
uniform electric field of magnitude 2 0 × 104 N C–1 [Fig |
1 | 491-494 | 7 An electron falls through a distance of 1 5 cm in a
uniform electric field of magnitude 2 0 × 104 N C–1 [Fig 1 |
1 | 492-495 | 5 cm in a
uniform electric field of magnitude 2 0 × 104 N C–1 [Fig 1 10(a)] |
1 | 493-496 | 0 × 104 N C–1 [Fig 1 10(a)] The
direction of the field is reversed keeping its magnitude unchanged
and a proton falls through the same distance [Fig |
1 | 494-497 | 1 10(a)] The
direction of the field is reversed keeping its magnitude unchanged
and a proton falls through the same distance [Fig 1 |
1 | 495-498 | 10(a)] The
direction of the field is reversed keeping its magnitude unchanged
and a proton falls through the same distance [Fig 1 10(b)] |
1 | 496-499 | The
direction of the field is reversed keeping its magnitude unchanged
and a proton falls through the same distance [Fig 1 10(b)] Compute
the time of fall in each case |
1 | 497-500 | 1 10(b)] Compute
the time of fall in each case Contrast the situation with that of ‘free
fall under gravity’ |
1 | 498-501 | 10(b)] Compute
the time of fall in each case Contrast the situation with that of ‘free
fall under gravity’ FIGURE 1 |
1 | 499-502 | Compute
the time of fall in each case Contrast the situation with that of ‘free
fall under gravity’ FIGURE 1 10
Solution In Fig |
1 | 500-503 | Contrast the situation with that of ‘free
fall under gravity’ FIGURE 1 10
Solution In Fig 1 |
1 | 501-504 | FIGURE 1 10
Solution In Fig 1 10(a) the field is upward, so the negatively charged
electron experiences a downward force of magnitude eE where E is
the magnitude of the electric field |
1 | 502-505 | 10
Solution In Fig 1 10(a) the field is upward, so the negatively charged
electron experiences a downward force of magnitude eE where E is
the magnitude of the electric field The acceleration of the electron is
ae = eE/me
where me is the mass of the electron |
1 | 503-506 | 1 10(a) the field is upward, so the negatively charged
electron experiences a downward force of magnitude eE where E is
the magnitude of the electric field The acceleration of the electron is
ae = eE/me
where me is the mass of the electron Starting from rest, the time required by the electron to fall through a
distance h is given by
2
2
e
e
e
h m
h
t
a
e E
=
=
For e = 1 |
1 | 504-507 | 10(a) the field is upward, so the negatively charged
electron experiences a downward force of magnitude eE where E is
the magnitude of the electric field The acceleration of the electron is
ae = eE/me
where me is the mass of the electron Starting from rest, the time required by the electron to fall through a
distance h is given by
2
2
e
e
e
h m
h
t
a
e E
=
=
For e = 1 6 × 10–19C, me = 9 |
1 | 505-508 | The acceleration of the electron is
ae = eE/me
where me is the mass of the electron Starting from rest, the time required by the electron to fall through a
distance h is given by
2
2
e
e
e
h m
h
t
a
e E
=
=
For e = 1 6 × 10–19C, me = 9 11 × 10–31 kg,
E = 2 |
1 | 506-509 | Starting from rest, the time required by the electron to fall through a
distance h is given by
2
2
e
e
e
h m
h
t
a
e E
=
=
For e = 1 6 × 10–19C, me = 9 11 × 10–31 kg,
E = 2 0 × 104 N C–1, h = 1 |
1 | 507-510 | 6 × 10–19C, me = 9 11 × 10–31 kg,
E = 2 0 × 104 N C–1, h = 1 5 × 10–2 m,
te = 2 |
1 | 508-511 | 11 × 10–31 kg,
E = 2 0 × 104 N C–1, h = 1 5 × 10–2 m,
te = 2 9 × 10–9s
In Fig |
1 | 509-512 | 0 × 104 N C–1, h = 1 5 × 10–2 m,
te = 2 9 × 10–9s
In Fig 1 |
1 | 510-513 | 5 × 10–2 m,
te = 2 9 × 10–9s
In Fig 1 10 (b), the field is downward, and the positively charged
proton experiences a downward force of magnitude eE |
1 | 511-514 | 9 × 10–9s
In Fig 1 10 (b), the field is downward, and the positively charged
proton experiences a downward force of magnitude eE The
acceleration of the proton is
ap = eE/mp
where mp is the mass of the proton; mp = 1 |
1 | 512-515 | 1 10 (b), the field is downward, and the positively charged
proton experiences a downward force of magnitude eE The
acceleration of the proton is
ap = eE/mp
where mp is the mass of the proton; mp = 1 67 × 10–27 kg |
1 | 513-516 | 10 (b), the field is downward, and the positively charged
proton experiences a downward force of magnitude eE The
acceleration of the proton is
ap = eE/mp
where mp is the mass of the proton; mp = 1 67 × 10–27 kg The time of
fall for the proton is
EXAMPLE 1 |
1 | 514-517 | The
acceleration of the proton is
ap = eE/mp
where mp is the mass of the proton; mp = 1 67 × 10–27 kg The time of
fall for the proton is
EXAMPLE 1 7
Rationalised 2023-24
18
Physics
EXAMPLE 1 |
1 | 515-518 | 67 × 10–27 kg The time of
fall for the proton is
EXAMPLE 1 7
Rationalised 2023-24
18
Physics
EXAMPLE 1 8
EXAMPLE 1 |
1 | 516-519 | The time of
fall for the proton is
EXAMPLE 1 7
Rationalised 2023-24
18
Physics
EXAMPLE 1 8
EXAMPLE 1 7
–7
2
2
1 3 10
s
p
p
p
h m
h
t |
1 | 517-520 | 7
Rationalised 2023-24
18
Physics
EXAMPLE 1 8
EXAMPLE 1 7
–7
2
2
1 3 10
s
p
p
p
h m
h
t a
e E
=
=
=
×
Thus, the heavier particle (proton) takes a greater time to fall through
the same distance |
1 | 518-521 | 8
EXAMPLE 1 7
–7
2
2
1 3 10
s
p
p
p
h m
h
t a
e E
=
=
=
×
Thus, the heavier particle (proton) takes a greater time to fall through
the same distance This is in basic contrast to the situation of ‘free
fall under gravity’ where the time of fall is independent of the mass of
the body |
1 | 519-522 | 7
–7
2
2
1 3 10
s
p
p
p
h m
h
t a
e E
=
=
=
×
Thus, the heavier particle (proton) takes a greater time to fall through
the same distance This is in basic contrast to the situation of ‘free
fall under gravity’ where the time of fall is independent of the mass of
the body Note that in this example we have ignored the acceleration
due to gravity in calculating the time of fall |
1 | 520-523 | a
e E
=
=
=
×
Thus, the heavier particle (proton) takes a greater time to fall through
the same distance This is in basic contrast to the situation of ‘free
fall under gravity’ where the time of fall is independent of the mass of
the body Note that in this example we have ignored the acceleration
due to gravity in calculating the time of fall To see if this is justified,
let us calculate the acceleration of the proton in the given electric
field:
p
p
e E
a
m
=
19
4
1
27
(1 6
10
C)
(2 0
10
N C
)
1 67
10
kg |
1 | 521-524 | This is in basic contrast to the situation of ‘free
fall under gravity’ where the time of fall is independent of the mass of
the body Note that in this example we have ignored the acceleration
due to gravity in calculating the time of fall To see if this is justified,
let us calculate the acceleration of the proton in the given electric
field:
p
p
e E
a
m
=
19
4
1
27
(1 6
10
C)
(2 0
10
N C
)
1 67
10
kg −
−
−
×
×
×
=
×
12
–2
1 9
10
m s |
1 | 522-525 | Note that in this example we have ignored the acceleration
due to gravity in calculating the time of fall To see if this is justified,
let us calculate the acceleration of the proton in the given electric
field:
p
p
e E
a
m
=
19
4
1
27
(1 6
10
C)
(2 0
10
N C
)
1 67
10
kg −
−
−
×
×
×
=
×
12
–2
1 9
10
m s =
×
which is enormous compared to the value of g (9 |
1 | 523-526 | To see if this is justified,
let us calculate the acceleration of the proton in the given electric
field:
p
p
e E
a
m
=
19
4
1
27
(1 6
10
C)
(2 0
10
N C
)
1 67
10
kg −
−
−
×
×
×
=
×
12
–2
1 9
10
m s =
×
which is enormous compared to the value of g (9 8 m s–2), the
acceleration due to gravity |
1 | 524-527 | −
−
−
×
×
×
=
×
12
–2
1 9
10
m s =
×
which is enormous compared to the value of g (9 8 m s–2), the
acceleration due to gravity The acceleration of the electron is even
greater |
1 | 525-528 | =
×
which is enormous compared to the value of g (9 8 m s–2), the
acceleration due to gravity The acceleration of the electron is even
greater Thus, the effect of acceleration due to gravity can be ignored
in this example |
1 | 526-529 | 8 m s–2), the
acceleration due to gravity The acceleration of the electron is even
greater Thus, the effect of acceleration due to gravity can be ignored
in this example Example 1 |
1 | 527-530 | The acceleration of the electron is even
greater Thus, the effect of acceleration due to gravity can be ignored
in this example Example 1 8 Two point charges q1 and q2, of magnitude +10–8 C and
–10–8 C, respectively, are placed 0 |
1 | 528-531 | Thus, the effect of acceleration due to gravity can be ignored
in this example Example 1 8 Two point charges q1 and q2, of magnitude +10–8 C and
–10–8 C, respectively, are placed 0 1 m apart |
1 | 529-532 | Example 1 8 Two point charges q1 and q2, of magnitude +10–8 C and
–10–8 C, respectively, are placed 0 1 m apart Calculate the electric
fields at points A, B and C shown in Fig |
1 | 530-533 | 8 Two point charges q1 and q2, of magnitude +10–8 C and
–10–8 C, respectively, are placed 0 1 m apart Calculate the electric
fields at points A, B and C shown in Fig 1 |
1 | 531-534 | 1 m apart Calculate the electric
fields at points A, B and C shown in Fig 1 11 |
1 | 532-535 | Calculate the electric
fields at points A, B and C shown in Fig 1 11 FIGURE 1 |
1 | 533-536 | 1 11 FIGURE 1 11
Solution The electric field vector E1A at A due to the positive charge
q1 points towards the right and has a magnitude
9
2
-2
8
1A
2
(9
10 Nm C )
(10
C)
(0 |
1 | 534-537 | 11 FIGURE 1 11
Solution The electric field vector E1A at A due to the positive charge
q1 points towards the right and has a magnitude
9
2
-2
8
1A
2
(9
10 Nm C )
(10
C)
(0 05m)
E
−
×
×
=
= 3 |
1 | 535-538 | FIGURE 1 11
Solution The electric field vector E1A at A due to the positive charge
q1 points towards the right and has a magnitude
9
2
-2
8
1A
2
(9
10 Nm C )
(10
C)
(0 05m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2A at A due to the negative charge q2 points
towards the right and has the same magnitude |
1 | 536-539 | 11
Solution The electric field vector E1A at A due to the positive charge
q1 points towards the right and has a magnitude
9
2
-2
8
1A
2
(9
10 Nm C )
(10
C)
(0 05m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2A at A due to the negative charge q2 points
towards the right and has the same magnitude Hence the magnitude
of the total electric field EA at A is
EA = E1A + E2A = 7 |
1 | 537-540 | 05m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2A at A due to the negative charge q2 points
towards the right and has the same magnitude Hence the magnitude
of the total electric field EA at A is
EA = E1A + E2A = 7 2 × 104 N C–1
EA is directed toward the right |
1 | 538-541 | 6 × 104 N C–1
The electric field vector E2A at A due to the negative charge q2 points
towards the right and has the same magnitude Hence the magnitude
of the total electric field EA at A is
EA = E1A + E2A = 7 2 × 104 N C–1
EA is directed toward the right Rationalised 2023-24
Electric Charges
and Fields
19
The electric field vector E1B at B due to the positive charge q1 points
towards the left and has a magnitude
9
2
–2
8
1B
2
(9
10 Nm C
)
(10
C)
(0 |
1 | 539-542 | Hence the magnitude
of the total electric field EA at A is
EA = E1A + E2A = 7 2 × 104 N C–1
EA is directed toward the right Rationalised 2023-24
Electric Charges
and Fields
19
The electric field vector E1B at B due to the positive charge q1 points
towards the left and has a magnitude
9
2
–2
8
1B
2
(9
10 Nm C
)
(10
C)
(0 05 m)
E
−
×
×
=
= 3 |
1 | 540-543 | 2 × 104 N C–1
EA is directed toward the right Rationalised 2023-24
Electric Charges
and Fields
19
The electric field vector E1B at B due to the positive charge q1 points
towards the left and has a magnitude
9
2
–2
8
1B
2
(9
10 Nm C
)
(10
C)
(0 05 m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2B at B due to the negative charge q2 points
towards the right and has a magnitude
9
2
–2
8
2B
2
(9
10 Nm C
)
(10
C)
(0 |
1 | 541-544 | Rationalised 2023-24
Electric Charges
and Fields
19
The electric field vector E1B at B due to the positive charge q1 points
towards the left and has a magnitude
9
2
–2
8
1B
2
(9
10 Nm C
)
(10
C)
(0 05 m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2B at B due to the negative charge q2 points
towards the right and has a magnitude
9
2
–2
8
2B
2
(9
10 Nm C
)
(10
C)
(0 15 m)
E
−
×
×
=
= 4 × 103 N C–1
The magnitude of the total electric field at B is
EB = E1B – E2B = 3 |
1 | 542-545 | 05 m)
E
−
×
×
=
= 3 6 × 104 N C–1
The electric field vector E2B at B due to the negative charge q2 points
towards the right and has a magnitude
9
2
–2
8
2B
2
(9
10 Nm C
)
(10
C)
(0 15 m)
E
−
×
×
=
= 4 × 103 N C–1
The magnitude of the total electric field at B is
EB = E1B – E2B = 3 2 × 104 N C–1
EB is directed towards the left |
1 | 543-546 | 6 × 104 N C–1
The electric field vector E2B at B due to the negative charge q2 points
towards the right and has a magnitude
9
2
–2
8
2B
2
(9
10 Nm C
)
(10
C)
(0 15 m)
E
−
×
×
=
= 4 × 103 N C–1
The magnitude of the total electric field at B is
EB = E1B – E2B = 3 2 × 104 N C–1
EB is directed towards the left The magnitude of each electric field vector at point C, due to charge
q1 and q2 is
9
2
–2
8
1C
2C
2
(9
10 Nm C
)
(10
C)
(0 |
1 | 544-547 | 15 m)
E
−
×
×
=
= 4 × 103 N C–1
The magnitude of the total electric field at B is
EB = E1B – E2B = 3 2 × 104 N C–1
EB is directed towards the left The magnitude of each electric field vector at point C, due to charge
q1 and q2 is
9
2
–2
8
1C
2C
2
(9
10 Nm C
)
(10
C)
(0 10 m)
E
E
−
×
×
=
=
= 9 × 103 N C–1
The directions in which these two vectors point are indicated in
Fig |
1 | 545-548 | 2 × 104 N C–1
EB is directed towards the left The magnitude of each electric field vector at point C, due to charge
q1 and q2 is
9
2
–2
8
1C
2C
2
(9
10 Nm C
)
(10
C)
(0 10 m)
E
E
−
×
×
=
=
= 9 × 103 N C–1
The directions in which these two vectors point are indicated in
Fig 1 |
1 | 546-549 | The magnitude of each electric field vector at point C, due to charge
q1 and q2 is
9
2
–2
8
1C
2C
2
(9
10 Nm C
)
(10
C)
(0 10 m)
E
E
−
×
×
=
=
= 9 × 103 N C–1
The directions in which these two vectors point are indicated in
Fig 1 11 |
1 | 547-550 | 10 m)
E
E
−
×
×
=
=
= 9 × 103 N C–1
The directions in which these two vectors point are indicated in
Fig 1 11 The resultant of these two vectors is
1
2
cos
cos
3
3
π
π
=
+
C
c
c
E
E
E
= 9 × 103 N C–1
EC points towards the right |
1 | 548-551 | 1 11 The resultant of these two vectors is
1
2
cos
cos
3
3
π
π
=
+
C
c
c
E
E
E
= 9 × 103 N C–1
EC points towards the right 1 |
1 | 549-552 | 11 The resultant of these two vectors is
1
2
cos
cos
3
3
π
π
=
+
C
c
c
E
E
E
= 9 × 103 N C–1
EC points towards the right 1 8 ELECTRIC FIELD LINES
We have studied electric field in the last section |
1 | 550-553 | The resultant of these two vectors is
1
2
cos
cos
3
3
π
π
=
+
C
c
c
E
E
E
= 9 × 103 N C–1
EC points towards the right 1 8 ELECTRIC FIELD LINES
We have studied electric field in the last section It is a vector quantity
and can be represented as we represent vectors |
1 | 551-554 | 1 8 ELECTRIC FIELD LINES
We have studied electric field in the last section It is a vector quantity
and can be represented as we represent vectors Let us try to represent E
due to a point charge pictorially |
1 | 552-555 | 8 ELECTRIC FIELD LINES
We have studied electric field in the last section It is a vector quantity
and can be represented as we represent vectors Let us try to represent E
due to a point charge pictorially Let the point charge be placed at the
origin |
1 | 553-556 | It is a vector quantity
and can be represented as we represent vectors Let us try to represent E
due to a point charge pictorially Let the point charge be placed at the
origin Draw vectors pointing along the direction of the
electric field with their lengths proportional to the strength
of the field at each point |
1 | 554-557 | Let us try to represent E
due to a point charge pictorially Let the point charge be placed at the
origin Draw vectors pointing along the direction of the
electric field with their lengths proportional to the strength
of the field at each point Since the magnitude of electric
field at a point decreases inversely as the square of the
distance of that point from the charge, the vector gets
shorter as one goes away from the origin, always pointing
radially outward |
1 | 555-558 | Let the point charge be placed at the
origin Draw vectors pointing along the direction of the
electric field with their lengths proportional to the strength
of the field at each point Since the magnitude of electric
field at a point decreases inversely as the square of the
distance of that point from the charge, the vector gets
shorter as one goes away from the origin, always pointing
radially outward Figure 1 |
1 | 556-559 | Draw vectors pointing along the direction of the
electric field with their lengths proportional to the strength
of the field at each point Since the magnitude of electric
field at a point decreases inversely as the square of the
distance of that point from the charge, the vector gets
shorter as one goes away from the origin, always pointing
radially outward Figure 1 12 shows such a picture |
1 | 557-560 | Since the magnitude of electric
field at a point decreases inversely as the square of the
distance of that point from the charge, the vector gets
shorter as one goes away from the origin, always pointing
radially outward Figure 1 12 shows such a picture In
this figure, each arrow indicates the electric field, i |
1 | 558-561 | Figure 1 12 shows such a picture In
this figure, each arrow indicates the electric field, i e |
1 | 559-562 | 12 shows such a picture In
this figure, each arrow indicates the electric field, i e , the
force acting on a unit positive charge, placed at the tail of
that arrow |
1 | 560-563 | In
this figure, each arrow indicates the electric field, i e , the
force acting on a unit positive charge, placed at the tail of
that arrow Connect the arrows pointing in one direction
and the resulting figure represents a field line |
1 | 561-564 | e , the
force acting on a unit positive charge, placed at the tail of
that arrow Connect the arrows pointing in one direction
and the resulting figure represents a field line We thus
get many field lines, all pointing outwards from the point
charge |
1 | 562-565 | , the
force acting on a unit positive charge, placed at the tail of
that arrow Connect the arrows pointing in one direction
and the resulting figure represents a field line We thus
get many field lines, all pointing outwards from the point
charge Have we lost the information about the strength
or magnitude of the field now, because it was contained
in the length of the arrow |
1 | 563-566 | Connect the arrows pointing in one direction
and the resulting figure represents a field line We thus
get many field lines, all pointing outwards from the point
charge Have we lost the information about the strength
or magnitude of the field now, because it was contained
in the length of the arrow No |
1 | 564-567 | We thus
get many field lines, all pointing outwards from the point
charge Have we lost the information about the strength
or magnitude of the field now, because it was contained
in the length of the arrow No Now the magnitude of the
field is represented by the density of field lines |
1 | 565-568 | Have we lost the information about the strength
or magnitude of the field now, because it was contained
in the length of the arrow No Now the magnitude of the
field is represented by the density of field lines E is strong
near the charge, so the density of field lines is more near
the charge and the lines are closer |
1 | 566-569 | No Now the magnitude of the
field is represented by the density of field lines E is strong
near the charge, so the density of field lines is more near
the charge and the lines are closer Away from the charge,
the field gets weaker and the density of field lines is less,
resulting in well-separated lines |
1 | 567-570 | Now the magnitude of the
field is represented by the density of field lines E is strong
near the charge, so the density of field lines is more near
the charge and the lines are closer Away from the charge,
the field gets weaker and the density of field lines is less,
resulting in well-separated lines Another person may draw more lines |
1 | 568-571 | E is strong
near the charge, so the density of field lines is more near
the charge and the lines are closer Away from the charge,
the field gets weaker and the density of field lines is less,
resulting in well-separated lines Another person may draw more lines But the number of lines is not
important |
1 | 569-572 | Away from the charge,
the field gets weaker and the density of field lines is less,
resulting in well-separated lines Another person may draw more lines But the number of lines is not
important In fact, an infinite number of lines can be drawn in any region |
1 | 570-573 | Another person may draw more lines But the number of lines is not
important In fact, an infinite number of lines can be drawn in any region FIGURE 1 |
1 | 571-574 | But the number of lines is not
important In fact, an infinite number of lines can be drawn in any region FIGURE 1 12 Field of a point charge |
1 | 572-575 | In fact, an infinite number of lines can be drawn in any region FIGURE 1 12 Field of a point charge EXAMPLE 1 |
1 | 573-576 | FIGURE 1 12 Field of a point charge EXAMPLE 1 8
Rationalised 2023-24
20
Physics
It is the relative density of lines in different regions which is
important |
1 | 574-577 | 12 Field of a point charge EXAMPLE 1 8
Rationalised 2023-24
20
Physics
It is the relative density of lines in different regions which is
important We draw the figure on the plane of paper, i |
1 | 575-578 | EXAMPLE 1 8
Rationalised 2023-24
20
Physics
It is the relative density of lines in different regions which is
important We draw the figure on the plane of paper, i e |
1 | 576-579 | 8
Rationalised 2023-24
20
Physics
It is the relative density of lines in different regions which is
important We draw the figure on the plane of paper, i e , in two-
dimensions but we live in three-dimensions |
1 | 577-580 | We draw the figure on the plane of paper, i e , in two-
dimensions but we live in three-dimensions So if one wishes
to estimate the density of field lines, one has to consider the
number of lines per unit cross-sectional area, perpendicular
to the lines |
1 | 578-581 | e , in two-
dimensions but we live in three-dimensions So if one wishes
to estimate the density of field lines, one has to consider the
number of lines per unit cross-sectional area, perpendicular
to the lines Since the electric field decreases as the square of
the distance from a point charge and the area enclosing the
charge increases as the square of the distance, the number
of field lines crossing the enclosing area remains constant,
whatever may be the distance of the area from the charge |
1 | 579-582 | , in two-
dimensions but we live in three-dimensions So if one wishes
to estimate the density of field lines, one has to consider the
number of lines per unit cross-sectional area, perpendicular
to the lines Since the electric field decreases as the square of
the distance from a point charge and the area enclosing the
charge increases as the square of the distance, the number
of field lines crossing the enclosing area remains constant,
whatever may be the distance of the area from the charge We started by saying that the field lines carry information
about the direction of electric field at different points in space |
1 | 580-583 | So if one wishes
to estimate the density of field lines, one has to consider the
number of lines per unit cross-sectional area, perpendicular
to the lines Since the electric field decreases as the square of
the distance from a point charge and the area enclosing the
charge increases as the square of the distance, the number
of field lines crossing the enclosing area remains constant,
whatever may be the distance of the area from the charge We started by saying that the field lines carry information
about the direction of electric field at different points in space Having drawn a certain set of field lines, the relative density
(i |
1 | 581-584 | Since the electric field decreases as the square of
the distance from a point charge and the area enclosing the
charge increases as the square of the distance, the number
of field lines crossing the enclosing area remains constant,
whatever may be the distance of the area from the charge We started by saying that the field lines carry information
about the direction of electric field at different points in space Having drawn a certain set of field lines, the relative density
(i e |
1 | 582-585 | We started by saying that the field lines carry information
about the direction of electric field at different points in space Having drawn a certain set of field lines, the relative density
(i e , closeness) of the field lines at different points indicates
the relative strength of electric field at those points |
1 | 583-586 | Having drawn a certain set of field lines, the relative density
(i e , closeness) of the field lines at different points indicates
the relative strength of electric field at those points The field
lines crowd where the field is strong and are spaced apart
where it is weak |
1 | 584-587 | e , closeness) of the field lines at different points indicates
the relative strength of electric field at those points The field
lines crowd where the field is strong and are spaced apart
where it is weak Figure 1 |
1 | 585-588 | , closeness) of the field lines at different points indicates
the relative strength of electric field at those points The field
lines crowd where the field is strong and are spaced apart
where it is weak Figure 1 13 shows a set of field lines |
1 | 586-589 | The field
lines crowd where the field is strong and are spaced apart
where it is weak Figure 1 13 shows a set of field lines We
can imagine two equal and small elements of area placed at points R and
S normal to the field lines there |
1 | 587-590 | Figure 1 13 shows a set of field lines We
can imagine two equal and small elements of area placed at points R and
S normal to the field lines there The number of field lines in our picture
cutting the area elements is proportional to the magnitude of field at
these points |
1 | 588-591 | 13 shows a set of field lines We
can imagine two equal and small elements of area placed at points R and
S normal to the field lines there The number of field lines in our picture
cutting the area elements is proportional to the magnitude of field at
these points The picture shows that the field at R is stronger than at S |
1 | 589-592 | We
can imagine two equal and small elements of area placed at points R and
S normal to the field lines there The number of field lines in our picture
cutting the area elements is proportional to the magnitude of field at
these points The picture shows that the field at R is stronger than at S To understand the dependence of the field lines on the area, or rather
the solid angle subtended by an area element, let us try to relate the
area with the solid angle, a generalisation of angle to three dimensions |
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