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1 | 1490-1493 | Any arbitrary infinitesimal
path can be resolved into two perpendicular displacements: One along
r and another perpendicular to r The work done corresponding to
the later will be zero 2 4 POTENTIAL DUE TO AN ELECTRIC DIPOLE
As we learnt in the last chapter, an electric dipole consists of two charges
q and –q separated by a (small) distance 2a |
1 | 1491-1494 | The work done corresponding to
the later will be zero 2 4 POTENTIAL DUE TO AN ELECTRIC DIPOLE
As we learnt in the last chapter, an electric dipole consists of two charges
q and –q separated by a (small) distance 2a Its total charge is zero |
1 | 1492-1495 | 2 4 POTENTIAL DUE TO AN ELECTRIC DIPOLE
As we learnt in the last chapter, an electric dipole consists of two charges
q and –q separated by a (small) distance 2a Its total charge is zero It is
characterised by a dipole moment vector p whose magnitude is q × 2a
and which points in the direction from –q to q (Fig |
1 | 1493-1496 | 4 POTENTIAL DUE TO AN ELECTRIC DIPOLE
As we learnt in the last chapter, an electric dipole consists of two charges
q and –q separated by a (small) distance 2a Its total charge is zero It is
characterised by a dipole moment vector p whose magnitude is q × 2a
and which points in the direction from –q to q (Fig 2 |
1 | 1494-1497 | Its total charge is zero It is
characterised by a dipole moment vector p whose magnitude is q × 2a
and which points in the direction from –q to q (Fig 2 5) |
1 | 1495-1498 | It is
characterised by a dipole moment vector p whose magnitude is q × 2a
and which points in the direction from –q to q (Fig 2 5) We also saw that
the electric field of a dipole at a point with position vector r depends not
just on the magnitude r, but also on the angle between r and p |
1 | 1496-1499 | 2 5) We also saw that
the electric field of a dipole at a point with position vector r depends not
just on the magnitude r, but also on the angle between r and p Further,
FIGURE 2 |
1 | 1497-1500 | 5) We also saw that
the electric field of a dipole at a point with position vector r depends not
just on the magnitude r, but also on the angle between r and p Further,
FIGURE 2 4 Variation of potential V with r [in units of
(Q/4pe0) m-1] (blue curve) and field with r [in units
of (Q/4pe0) m-2] (black curve) for a point charge Q |
1 | 1498-1501 | We also saw that
the electric field of a dipole at a point with position vector r depends not
just on the magnitude r, but also on the angle between r and p Further,
FIGURE 2 4 Variation of potential V with r [in units of
(Q/4pe0) m-1] (blue curve) and field with r [in units
of (Q/4pe0) m-2] (black curve) for a point charge Q Rationalised 2023-24
Physics
50
the field falls off, at large distance, not as
1/r 2 (typical of field due to a single charge)
but as 1/r3 |
1 | 1499-1502 | Further,
FIGURE 2 4 Variation of potential V with r [in units of
(Q/4pe0) m-1] (blue curve) and field with r [in units
of (Q/4pe0) m-2] (black curve) for a point charge Q Rationalised 2023-24
Physics
50
the field falls off, at large distance, not as
1/r 2 (typical of field due to a single charge)
but as 1/r3 We, now, determine the electric
potential due to a dipole and contrast it
with the potential due to a single charge |
1 | 1500-1503 | 4 Variation of potential V with r [in units of
(Q/4pe0) m-1] (blue curve) and field with r [in units
of (Q/4pe0) m-2] (black curve) for a point charge Q Rationalised 2023-24
Physics
50
the field falls off, at large distance, not as
1/r 2 (typical of field due to a single charge)
but as 1/r3 We, now, determine the electric
potential due to a dipole and contrast it
with the potential due to a single charge As before, we take the origin at the
centre of the dipole |
1 | 1501-1504 | Rationalised 2023-24
Physics
50
the field falls off, at large distance, not as
1/r 2 (typical of field due to a single charge)
but as 1/r3 We, now, determine the electric
potential due to a dipole and contrast it
with the potential due to a single charge As before, we take the origin at the
centre of the dipole Now we know that the
electric field obeys the superposition
principle |
1 | 1502-1505 | We, now, determine the electric
potential due to a dipole and contrast it
with the potential due to a single charge As before, we take the origin at the
centre of the dipole Now we know that the
electric field obeys the superposition
principle Since potential is related to the
work done by the field, electrostatic
potential also follows the superposition
principle |
1 | 1503-1506 | As before, we take the origin at the
centre of the dipole Now we know that the
electric field obeys the superposition
principle Since potential is related to the
work done by the field, electrostatic
potential also follows the superposition
principle Thus, the potential due to the
dipole is the sum of potentials due to the
charges q and –q
V
rq
rq
=
−
41
0
1
2
πε
(2 |
1 | 1504-1507 | Now we know that the
electric field obeys the superposition
principle Since potential is related to the
work done by the field, electrostatic
potential also follows the superposition
principle Thus, the potential due to the
dipole is the sum of potentials due to the
charges q and –q
V
rq
rq
=
−
41
0
1
2
πε
(2 9)
where r1 and r2 are the distances of the
point P from q and –q, respectively |
1 | 1505-1508 | Since potential is related to the
work done by the field, electrostatic
potential also follows the superposition
principle Thus, the potential due to the
dipole is the sum of potentials due to the
charges q and –q
V
rq
rq
=
−
41
0
1
2
πε
(2 9)
where r1 and r2 are the distances of the
point P from q and –q, respectively Now, by geometry,
2
2
2
1
2
r
r
a
ar
=
+
−
cosq
2
2
2
2
2
r
r
a
ar
=
+
+
cosq
(2 |
1 | 1506-1509 | Thus, the potential due to the
dipole is the sum of potentials due to the
charges q and –q
V
rq
rq
=
−
41
0
1
2
πε
(2 9)
where r1 and r2 are the distances of the
point P from q and –q, respectively Now, by geometry,
2
2
2
1
2
r
r
a
ar
=
+
−
cosq
2
2
2
2
2
r
r
a
ar
=
+
+
cosq
(2 10)
We take r much greater than a (
r a
) and retain terms only upto
the first order in a/r
r
r
a
r
ra
1
2
2
2
2
1
2
=
−
+
cosθ
≅
−
r
a
r
2
1
2
cosθ
(2 |
1 | 1507-1510 | 9)
where r1 and r2 are the distances of the
point P from q and –q, respectively Now, by geometry,
2
2
2
1
2
r
r
a
ar
=
+
−
cosq
2
2
2
2
2
r
r
a
ar
=
+
+
cosq
(2 10)
We take r much greater than a (
r a
) and retain terms only upto
the first order in a/r
r
r
a
r
ra
1
2
2
2
2
1
2
=
−
+
cosθ
≅
−
r
a
r
2
1
2
cosθ
(2 11)
Similarly,
r
r
a
r
2
2
2
1
2
≅
+
cosθ
(2 |
1 | 1508-1511 | Now, by geometry,
2
2
2
1
2
r
r
a
ar
=
+
−
cosq
2
2
2
2
2
r
r
a
ar
=
+
+
cosq
(2 10)
We take r much greater than a (
r a
) and retain terms only upto
the first order in a/r
r
r
a
r
ra
1
2
2
2
2
1
2
=
−
+
cosθ
≅
−
r
a
r
2
1
2
cosθ
(2 11)
Similarly,
r
r
a
r
2
2
2
1
2
≅
+
cosθ
(2 12)
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
1
1 1
2
1 1
1
1 2
r
r
a
r
r
ra
≅
−
≅
+
−
cos
cos
/
θ
θ
[2 |
1 | 1509-1512 | 10)
We take r much greater than a (
r a
) and retain terms only upto
the first order in a/r
r
r
a
r
ra
1
2
2
2
2
1
2
=
−
+
cosθ
≅
−
r
a
r
2
1
2
cosθ
(2 11)
Similarly,
r
r
a
r
2
2
2
1
2
≅
+
cosθ
(2 12)
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
1
1 1
2
1 1
1
1 2
r
r
a
r
r
ra
≅
−
≅
+
−
cos
cos
/
θ
θ
[2 13(a)]
1
1 1
2
1 1
2
1 2
r
r
a
r
r
ra
≅
+
≅
−
−
cos
cos
/
θ
θ
[2 |
1 | 1510-1513 | 11)
Similarly,
r
r
a
r
2
2
2
1
2
≅
+
cosθ
(2 12)
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
1
1 1
2
1 1
1
1 2
r
r
a
r
r
ra
≅
−
≅
+
−
cos
cos
/
θ
θ
[2 13(a)]
1
1 1
2
1 1
2
1 2
r
r
a
r
r
ra
≅
+
≅
−
−
cos
cos
/
θ
θ
[2 13(b)]
Using Eqs |
1 | 1511-1514 | 12)
Using the Binomial theorem and retaining terms upto the first order
in a/r ; we obtain,
1
1 1
2
1 1
1
1 2
r
r
a
r
r
ra
≅
−
≅
+
−
cos
cos
/
θ
θ
[2 13(a)]
1
1 1
2
1 1
2
1 2
r
r
a
r
r
ra
≅
+
≅
−
−
cos
cos
/
θ
θ
[2 13(b)]
Using Eqs (2 |
1 | 1512-1515 | 13(a)]
1
1 1
2
1 1
2
1 2
r
r
a
r
r
ra
≅
+
≅
−
−
cos
cos
/
θ
θ
[2 13(b)]
Using Eqs (2 9) and (2 |
1 | 1513-1516 | 13(b)]
Using Eqs (2 9) and (2 13) and p = 2qa, we get
V
q
a
r
p
r
=
=
4
4
0
2
0
2
π
π
ε
θ
θ
ε
2 cos
cos
(2 |
1 | 1514-1517 | (2 9) and (2 13) and p = 2qa, we get
V
q
a
r
p
r
=
=
4
4
0
2
0
2
π
π
ε
θ
θ
ε
2 cos
cos
(2 14)
Now, p cos q = p |
1 | 1515-1518 | 9) and (2 13) and p = 2qa, we get
V
q
a
r
p
r
=
=
4
4
0
2
0
2
π
π
ε
θ
θ
ε
2 cos
cos
(2 14)
Now, p cos q = p rˆ
FIGURE 2 |
1 | 1516-1519 | 13) and p = 2qa, we get
V
q
a
r
p
r
=
=
4
4
0
2
0
2
π
π
ε
θ
θ
ε
2 cos
cos
(2 14)
Now, p cos q = p rˆ
FIGURE 2 5 Quantities involved in the calculation
of potential due to a dipole |
1 | 1517-1520 | 14)
Now, p cos q = p rˆ
FIGURE 2 5 Quantities involved in the calculation
of potential due to a dipole Rationalised 2023-24
Electrostatic Potential
and Capacitance
51
where ˆr is the unit vector along the position vector OP |
1 | 1518-1521 | rˆ
FIGURE 2 5 Quantities involved in the calculation
of potential due to a dipole Rationalised 2023-24
Electrostatic Potential
and Capacitance
51
where ˆr is the unit vector along the position vector OP The electric potential of a dipole is then given by
V
r
=
41
0
2
πε
p |
1 | 1519-1522 | 5 Quantities involved in the calculation
of potential due to a dipole Rationalised 2023-24
Electrostatic Potential
and Capacitance
51
where ˆr is the unit vector along the position vector OP The electric potential of a dipole is then given by
V
r
=
41
0
2
πε
p rˆ
; (r >> a)
(2 |
1 | 1520-1523 | Rationalised 2023-24
Electrostatic Potential
and Capacitance
51
where ˆr is the unit vector along the position vector OP The electric potential of a dipole is then given by
V
r
=
41
0
2
πε
p rˆ
; (r >> a)
(2 15)
Equation (2 |
1 | 1521-1524 | The electric potential of a dipole is then given by
V
r
=
41
0
2
πε
p rˆ
; (r >> a)
(2 15)
Equation (2 15) is, as indicated, approximately true only for distances
large compared to the size of the dipole, so that higher order terms in
a/r are negligible |
1 | 1522-1525 | rˆ
; (r >> a)
(2 15)
Equation (2 15) is, as indicated, approximately true only for distances
large compared to the size of the dipole, so that higher order terms in
a/r are negligible For a point dipole p at the origin, Eq |
1 | 1523-1526 | 15)
Equation (2 15) is, as indicated, approximately true only for distances
large compared to the size of the dipole, so that higher order terms in
a/r are negligible For a point dipole p at the origin, Eq (2 |
1 | 1524-1527 | 15) is, as indicated, approximately true only for distances
large compared to the size of the dipole, so that higher order terms in
a/r are negligible For a point dipole p at the origin, Eq (2 15) is, however,
exact |
1 | 1525-1528 | For a point dipole p at the origin, Eq (2 15) is, however,
exact From Eq |
1 | 1526-1529 | (2 15) is, however,
exact From Eq (2 |
1 | 1527-1530 | 15) is, however,
exact From Eq (2 15), potential on the dipole axis (q = 0, p ) is given by
2
0
1
4
p
V
εr
= ±
π
(2 |
1 | 1528-1531 | From Eq (2 15), potential on the dipole axis (q = 0, p ) is given by
2
0
1
4
p
V
εr
= ±
π
(2 16)
(Positive sign for q = 0, negative sign for q = p |
1 | 1529-1532 | (2 15), potential on the dipole axis (q = 0, p ) is given by
2
0
1
4
p
V
εr
= ±
π
(2 16)
(Positive sign for q = 0, negative sign for q = p ) The potential in the
equatorial plane (q = p/2) is zero |
1 | 1530-1533 | 15), potential on the dipole axis (q = 0, p ) is given by
2
0
1
4
p
V
εr
= ±
π
(2 16)
(Positive sign for q = 0, negative sign for q = p ) The potential in the
equatorial plane (q = p/2) is zero The important contrasting features of electric potential of a dipole
from that due to a single charge are clear from Eqs |
1 | 1531-1534 | 16)
(Positive sign for q = 0, negative sign for q = p ) The potential in the
equatorial plane (q = p/2) is zero The important contrasting features of electric potential of a dipole
from that due to a single charge are clear from Eqs (2 |
1 | 1532-1535 | ) The potential in the
equatorial plane (q = p/2) is zero The important contrasting features of electric potential of a dipole
from that due to a single charge are clear from Eqs (2 8) and (2 |
1 | 1533-1536 | The important contrasting features of electric potential of a dipole
from that due to a single charge are clear from Eqs (2 8) and (2 15):
(i)
The potential due to a dipole depends not just on r but also on the
angle between the position vector r and the dipole moment vector p |
1 | 1534-1537 | (2 8) and (2 15):
(i)
The potential due to a dipole depends not just on r but also on the
angle between the position vector r and the dipole moment vector p (It is, however, axially symmetric about p |
1 | 1535-1538 | 8) and (2 15):
(i)
The potential due to a dipole depends not just on r but also on the
angle between the position vector r and the dipole moment vector p (It is, however, axially symmetric about p That is, if you rotate the
position vector r about p, keeping q fixed, the points corresponding
to P on the cone so generated will have the same potential as at P |
1 | 1536-1539 | 15):
(i)
The potential due to a dipole depends not just on r but also on the
angle between the position vector r and the dipole moment vector p (It is, however, axially symmetric about p That is, if you rotate the
position vector r about p, keeping q fixed, the points corresponding
to P on the cone so generated will have the same potential as at P )
(ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as
1/r, characteristic of the potential due to a single charge |
1 | 1537-1540 | (It is, however, axially symmetric about p That is, if you rotate the
position vector r about p, keeping q fixed, the points corresponding
to P on the cone so generated will have the same potential as at P )
(ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as
1/r, characteristic of the potential due to a single charge (You can
refer to the Fig |
1 | 1538-1541 | That is, if you rotate the
position vector r about p, keeping q fixed, the points corresponding
to P on the cone so generated will have the same potential as at P )
(ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as
1/r, characteristic of the potential due to a single charge (You can
refer to the Fig 2 |
1 | 1539-1542 | )
(ii) The electric dipole potential falls off, at large distance, as 1/r 2, not as
1/r, characteristic of the potential due to a single charge (You can
refer to the Fig 2 5 for graphs of 1/r 2 versus r and 1/r versus r,
drawn there in another context |
1 | 1540-1543 | (You can
refer to the Fig 2 5 for graphs of 1/r 2 versus r and 1/r versus r,
drawn there in another context )
2 |
1 | 1541-1544 | 2 5 for graphs of 1/r 2 versus r and 1/r versus r,
drawn there in another context )
2 5 POTENTIAL DUE TO A SYSTEM OF CHARGES
Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…,
rn relative to some origin (Fig |
1 | 1542-1545 | 5 for graphs of 1/r 2 versus r and 1/r versus r,
drawn there in another context )
2 5 POTENTIAL DUE TO A SYSTEM OF CHARGES
Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…,
rn relative to some origin (Fig 2 |
1 | 1543-1546 | )
2 5 POTENTIAL DUE TO A SYSTEM OF CHARGES
Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…,
rn relative to some origin (Fig 2 6) |
1 | 1544-1547 | 5 POTENTIAL DUE TO A SYSTEM OF CHARGES
Consider a system of charges q1, q2,…, qn with position vectors r1, r2,…,
rn relative to some origin (Fig 2 6) The potential V1 at P due to the charge
q1 is
1
1
0
1P
1
4
q
V
εr
=
π
where r1P is the distance between q1 and P |
1 | 1545-1548 | 2 6) The potential V1 at P due to the charge
q1 is
1
1
0
1P
1
4
q
V
εr
=
π
where r1P is the distance between q1 and P Similarly, the potential V2 at P due to q2 and
V3 due to q3 are given by
2
2
0
2P
1
4
q
V
εr
=
π
,
3
3
0
3P
1
4
q
V
εr
=
π
where r2P and r3P are the distances of P from
charges q2 and q3, respectively; and so on for the
potential due to other charges |
1 | 1546-1549 | 6) The potential V1 at P due to the charge
q1 is
1
1
0
1P
1
4
q
V
εr
=
π
where r1P is the distance between q1 and P Similarly, the potential V2 at P due to q2 and
V3 due to q3 are given by
2
2
0
2P
1
4
q
V
εr
=
π
,
3
3
0
3P
1
4
q
V
εr
=
π
where r2P and r3P are the distances of P from
charges q2 and q3, respectively; and so on for the
potential due to other charges By the
superposition principle, the potential V at P due
to the total charge configuration is the algebraic
sum of the potentials due to the individual
charges
V = V1 + V2 + |
1 | 1547-1550 | The potential V1 at P due to the charge
q1 is
1
1
0
1P
1
4
q
V
εr
=
π
where r1P is the distance between q1 and P Similarly, the potential V2 at P due to q2 and
V3 due to q3 are given by
2
2
0
2P
1
4
q
V
εr
=
π
,
3
3
0
3P
1
4
q
V
εr
=
π
where r2P and r3P are the distances of P from
charges q2 and q3, respectively; and so on for the
potential due to other charges By the
superposition principle, the potential V at P due
to the total charge configuration is the algebraic
sum of the potentials due to the individual
charges
V = V1 + V2 + + Vn
(2 |
1 | 1548-1551 | Similarly, the potential V2 at P due to q2 and
V3 due to q3 are given by
2
2
0
2P
1
4
q
V
εr
=
π
,
3
3
0
3P
1
4
q
V
εr
=
π
where r2P and r3P are the distances of P from
charges q2 and q3, respectively; and so on for the
potential due to other charges By the
superposition principle, the potential V at P due
to the total charge configuration is the algebraic
sum of the potentials due to the individual
charges
V = V1 + V2 + + Vn
(2 17)
FIGURE 2 |
1 | 1549-1552 | By the
superposition principle, the potential V at P due
to the total charge configuration is the algebraic
sum of the potentials due to the individual
charges
V = V1 + V2 + + Vn
(2 17)
FIGURE 2 6 Potential at a point due to a
system of charges is the sum of potentials
due to individual charges |
1 | 1550-1553 | + Vn
(2 17)
FIGURE 2 6 Potential at a point due to a
system of charges is the sum of potentials
due to individual charges Rationalised 2023-24
Physics
52
EXAMPLE 2 |
1 | 1551-1554 | 17)
FIGURE 2 6 Potential at a point due to a
system of charges is the sum of potentials
due to individual charges Rationalised 2023-24
Physics
52
EXAMPLE 2 2
=
+
+
+
41
0
1
1
2
2
πε
rq
rq
q
r
n
n
P
P
P |
1 | 1552-1555 | 6 Potential at a point due to a
system of charges is the sum of potentials
due to individual charges Rationalised 2023-24
Physics
52
EXAMPLE 2 2
=
+
+
+
41
0
1
1
2
2
πε
rq
rq
q
r
n
n
P
P
P (2 |
1 | 1553-1556 | Rationalised 2023-24
Physics
52
EXAMPLE 2 2
=
+
+
+
41
0
1
1
2
2
πε
rq
rq
q
r
n
n
P
P
P (2 18)
If we have a continuous charge distribution characterised by a charge
density r (r), we divide it, as before, into small volume elements each of
size Dv and carrying a charge rDv |
1 | 1554-1557 | 2
=
+
+
+
41
0
1
1
2
2
πε
rq
rq
q
r
n
n
P
P
P (2 18)
If we have a continuous charge distribution characterised by a charge
density r (r), we divide it, as before, into small volume elements each of
size Dv and carrying a charge rDv We then determine the potential due
to each volume element and sum (strictly speaking , integrate) over all
such contributions, and thus determine the potential due to the entire
distribution |
1 | 1555-1558 | (2 18)
If we have a continuous charge distribution characterised by a charge
density r (r), we divide it, as before, into small volume elements each of
size Dv and carrying a charge rDv We then determine the potential due
to each volume element and sum (strictly speaking , integrate) over all
such contributions, and thus determine the potential due to the entire
distribution We have seen in Chapter 1 that for a uniformly charged spherical shell,
the electric field outside the shell is as if the entire charge is concentrated
at the centre |
1 | 1556-1559 | 18)
If we have a continuous charge distribution characterised by a charge
density r (r), we divide it, as before, into small volume elements each of
size Dv and carrying a charge rDv We then determine the potential due
to each volume element and sum (strictly speaking , integrate) over all
such contributions, and thus determine the potential due to the entire
distribution We have seen in Chapter 1 that for a uniformly charged spherical shell,
the electric field outside the shell is as if the entire charge is concentrated
at the centre Thus, the potential outside the shell is given by
0
1
4
q
V
r
ε
=
π
(
)
r
≥R
[2 |
1 | 1557-1560 | We then determine the potential due
to each volume element and sum (strictly speaking , integrate) over all
such contributions, and thus determine the potential due to the entire
distribution We have seen in Chapter 1 that for a uniformly charged spherical shell,
the electric field outside the shell is as if the entire charge is concentrated
at the centre Thus, the potential outside the shell is given by
0
1
4
q
V
r
ε
=
π
(
)
r
≥R
[2 19(a)]
where q is the total charge on the shell and R its radius |
1 | 1558-1561 | We have seen in Chapter 1 that for a uniformly charged spherical shell,
the electric field outside the shell is as if the entire charge is concentrated
at the centre Thus, the potential outside the shell is given by
0
1
4
q
V
r
ε
=
π
(
)
r
≥R
[2 19(a)]
where q is the total charge on the shell and R its radius The electric field
inside the shell is zero |
1 | 1559-1562 | Thus, the potential outside the shell is given by
0
1
4
q
V
r
ε
=
π
(
)
r
≥R
[2 19(a)]
where q is the total charge on the shell and R its radius The electric field
inside the shell is zero This implies (Section 2 |
1 | 1560-1563 | 19(a)]
where q is the total charge on the shell and R its radius The electric field
inside the shell is zero This implies (Section 2 6) that potential is constant
inside the shell (as no work is done in moving a charge inside the shell),
and, therefore, equals its value at the surface, which is
0
1
4
q
V
R
ε
=
π
[2 |
1 | 1561-1564 | The electric field
inside the shell is zero This implies (Section 2 6) that potential is constant
inside the shell (as no work is done in moving a charge inside the shell),
and, therefore, equals its value at the surface, which is
0
1
4
q
V
R
ε
=
π
[2 19(b)]
Example 2 |
1 | 1562-1565 | This implies (Section 2 6) that potential is constant
inside the shell (as no work is done in moving a charge inside the shell),
and, therefore, equals its value at the surface, which is
0
1
4
q
V
R
ε
=
π
[2 19(b)]
Example 2 2 Two charges 3 × 10–8 C and –2 × 10–8 C are located
15 cm apart |
1 | 1563-1566 | 6) that potential is constant
inside the shell (as no work is done in moving a charge inside the shell),
and, therefore, equals its value at the surface, which is
0
1
4
q
V
R
ε
=
π
[2 19(b)]
Example 2 2 Two charges 3 × 10–8 C and –2 × 10–8 C are located
15 cm apart At what point on the line joining the two charges is the
electric potential zero |
1 | 1564-1567 | 19(b)]
Example 2 2 Two charges 3 × 10–8 C and –2 × 10–8 C are located
15 cm apart At what point on the line joining the two charges is the
electric potential zero Take the potential at infinity to be zero |
1 | 1565-1568 | 2 Two charges 3 × 10–8 C and –2 × 10–8 C are located
15 cm apart At what point on the line joining the two charges is the
electric potential zero Take the potential at infinity to be zero Solution Let us take the origin O at the location of the positive charge |
1 | 1566-1569 | At what point on the line joining the two charges is the
electric potential zero Take the potential at infinity to be zero Solution Let us take the origin O at the location of the positive charge The line joining the two charges is taken to be the x-axis; the negative
charge is taken to be on the right side of the origin (Fig |
1 | 1567-1570 | Take the potential at infinity to be zero Solution Let us take the origin O at the location of the positive charge The line joining the two charges is taken to be the x-axis; the negative
charge is taken to be on the right side of the origin (Fig 2 |
1 | 1568-1571 | Solution Let us take the origin O at the location of the positive charge The line joining the two charges is taken to be the x-axis; the negative
charge is taken to be on the right side of the origin (Fig 2 7) |
1 | 1569-1572 | The line joining the two charges is taken to be the x-axis; the negative
charge is taken to be on the right side of the origin (Fig 2 7) FIGURE 2 |
1 | 1570-1573 | 2 7) FIGURE 2 7
Let P be the required point on the x-axis where the potential is zero |
1 | 1571-1574 | 7) FIGURE 2 7
Let P be the required point on the x-axis where the potential is zero If x is the x-coordinate of P, obviously x must be positive |
1 | 1572-1575 | FIGURE 2 7
Let P be the required point on the x-axis where the potential is zero If x is the x-coordinate of P, obviously x must be positive (There is no
possibility of potentials due to the two charges adding up to zero for
x < 0 |
1 | 1573-1576 | 7
Let P be the required point on the x-axis where the potential is zero If x is the x-coordinate of P, obviously x must be positive (There is no
possibility of potentials due to the two charges adding up to zero for
x < 0 ) If x lies between O and A, we have
41
3 10
10
152 10
10
0
0
8
2
8
2
πε
××
−
−×
×
=
–
–
–
–
(
)
x
x
where x is in cm |
1 | 1574-1577 | If x is the x-coordinate of P, obviously x must be positive (There is no
possibility of potentials due to the two charges adding up to zero for
x < 0 ) If x lies between O and A, we have
41
3 10
10
152 10
10
0
0
8
2
8
2
πε
××
−
−×
×
=
–
–
–
–
(
)
x
x
where x is in cm That is,
3
2
0
15
x
x
−
=
−
which gives x = 9 cm |
1 | 1575-1578 | (There is no
possibility of potentials due to the two charges adding up to zero for
x < 0 ) If x lies between O and A, we have
41
3 10
10
152 10
10
0
0
8
2
8
2
πε
××
−
−×
×
=
–
–
–
–
(
)
x
x
where x is in cm That is,
3
2
0
15
x
x
−
=
−
which gives x = 9 cm If x lies on the extended line OA, the required condition is
3
2
0
15
x
−x
=
−
Rationalised 2023-24
Electrostatic Potential
and Capacitance
53
EXAMPLE 2 |
1 | 1576-1579 | ) If x lies between O and A, we have
41
3 10
10
152 10
10
0
0
8
2
8
2
πε
××
−
−×
×
=
–
–
–
–
(
)
x
x
where x is in cm That is,
3
2
0
15
x
x
−
=
−
which gives x = 9 cm If x lies on the extended line OA, the required condition is
3
2
0
15
x
−x
=
−
Rationalised 2023-24
Electrostatic Potential
and Capacitance
53
EXAMPLE 2 2
which gives
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the
positive charge on the side of the negative charge |
1 | 1577-1580 | That is,
3
2
0
15
x
x
−
=
−
which gives x = 9 cm If x lies on the extended line OA, the required condition is
3
2
0
15
x
−x
=
−
Rationalised 2023-24
Electrostatic Potential
and Capacitance
53
EXAMPLE 2 2
which gives
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the
positive charge on the side of the negative charge Note that the
formula for potential used in the calculation required choosing
potential to be zero at infinity |
1 | 1578-1581 | If x lies on the extended line OA, the required condition is
3
2
0
15
x
−x
=
−
Rationalised 2023-24
Electrostatic Potential
and Capacitance
53
EXAMPLE 2 2
which gives
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the
positive charge on the side of the negative charge Note that the
formula for potential used in the calculation required choosing
potential to be zero at infinity Example 2 |
1 | 1579-1582 | 2
which gives
x = 45 cm
Thus, electric potential is zero at 9 cm and 45 cm away from the
positive charge on the side of the negative charge Note that the
formula for potential used in the calculation required choosing
potential to be zero at infinity Example 2 3 Figures 2 |
1 | 1580-1583 | Note that the
formula for potential used in the calculation required choosing
potential to be zero at infinity Example 2 3 Figures 2 8 (a) and (b) show the field lines of a positive
and negative point charge respectively |
1 | 1581-1584 | Example 2 3 Figures 2 8 (a) and (b) show the field lines of a positive
and negative point charge respectively FIGURE 2 |
1 | 1582-1585 | 3 Figures 2 8 (a) and (b) show the field lines of a positive
and negative point charge respectively FIGURE 2 8
(a) Give the signs of the potential difference VP – VQ; VB – VA |
1 | 1583-1586 | 8 (a) and (b) show the field lines of a positive
and negative point charge respectively FIGURE 2 8
(a) Give the signs of the potential difference VP – VQ; VB – VA (b) Give the sign of the potential energy difference of a small negative
charge between the points Q and P; A and B |
1 | 1584-1587 | FIGURE 2 8
(a) Give the signs of the potential difference VP – VQ; VB – VA (b) Give the sign of the potential energy difference of a small negative
charge between the points Q and P; A and B (c) Give the sign of the work done by the field in moving a small
positive charge from Q to P |
1 | 1585-1588 | 8
(a) Give the signs of the potential difference VP – VQ; VB – VA (b) Give the sign of the potential energy difference of a small negative
charge between the points Q and P; A and B (c) Give the sign of the work done by the field in moving a small
positive charge from Q to P (d) Give the sign of the work done by the external agency in moving
a small negative charge from B to A |
1 | 1586-1589 | (b) Give the sign of the potential energy difference of a small negative
charge between the points Q and P; A and B (c) Give the sign of the work done by the field in moving a small
positive charge from Q to P (d) Give the sign of the work done by the external agency in moving
a small negative charge from B to A (e) Does the kinetic energy of a small negative charge increase or
decrease in going from B to A |
1 | 1587-1590 | (c) Give the sign of the work done by the field in moving a small
positive charge from Q to P (d) Give the sign of the work done by the external agency in moving
a small negative charge from B to A (e) Does the kinetic energy of a small negative charge increase or
decrease in going from B to A Solution
(a) As
1
V
∝r
, VP > VQ |
1 | 1588-1591 | (d) Give the sign of the work done by the external agency in moving
a small negative charge from B to A (e) Does the kinetic energy of a small negative charge increase or
decrease in going from B to A Solution
(a) As
1
V
∝r
, VP > VQ Thus, (VP – VQ) is positive |
1 | 1589-1592 | (e) Does the kinetic energy of a small negative charge increase or
decrease in going from B to A Solution
(a) As
1
V
∝r
, VP > VQ Thus, (VP – VQ) is positive Also VB is less negative
than VA |
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