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1 | 1590-1593 | Solution
(a) As
1
V
∝r
, VP > VQ Thus, (VP – VQ) is positive Also VB is less negative
than VA Thus, VB > VA or (VB – VA) is positive |
1 | 1591-1594 | Thus, (VP – VQ) is positive Also VB is less negative
than VA Thus, VB > VA or (VB – VA) is positive (b) A small negative charge will be attracted towards positive charge |
1 | 1592-1595 | Also VB is less negative
than VA Thus, VB > VA or (VB – VA) is positive (b) A small negative charge will be attracted towards positive charge The negative charge moves from higher potential energy to lower
potential energy |
1 | 1593-1596 | Thus, VB > VA or (VB – VA) is positive (b) A small negative charge will be attracted towards positive charge The negative charge moves from higher potential energy to lower
potential energy Therefore the sign of potential energy difference
of a small negative charge between Q and P is positive |
1 | 1594-1597 | (b) A small negative charge will be attracted towards positive charge The negative charge moves from higher potential energy to lower
potential energy Therefore the sign of potential energy difference
of a small negative charge between Q and P is positive Similarly, (P |
1 | 1595-1598 | The negative charge moves from higher potential energy to lower
potential energy Therefore the sign of potential energy difference
of a small negative charge between Q and P is positive Similarly, (P E |
1 | 1596-1599 | Therefore the sign of potential energy difference
of a small negative charge between Q and P is positive Similarly, (P E )A > (P |
1 | 1597-1600 | Similarly, (P E )A > (P E |
1 | 1598-1601 | E )A > (P E )B
and hence sign of potential energy
differences is positive |
1 | 1599-1602 | )A > (P E )B
and hence sign of potential energy
differences is positive (c) In moving a small positive charge from Q to P, work has to be
done by an external agency against the electric field |
1 | 1600-1603 | E )B
and hence sign of potential energy
differences is positive (c) In moving a small positive charge from Q to P, work has to be
done by an external agency against the electric field Therefore,
work done by the field is negative |
1 | 1601-1604 | )B
and hence sign of potential energy
differences is positive (c) In moving a small positive charge from Q to P, work has to be
done by an external agency against the electric field Therefore,
work done by the field is negative (d) In moving a small negative charge from B to A work has to be
done by the external agency |
1 | 1602-1605 | (c) In moving a small positive charge from Q to P, work has to be
done by an external agency against the electric field Therefore,
work done by the field is negative (d) In moving a small negative charge from B to A work has to be
done by the external agency It is positive |
1 | 1603-1606 | Therefore,
work done by the field is negative (d) In moving a small negative charge from B to A work has to be
done by the external agency It is positive (e) Due to force of repulsion on the negative charge, velocity decreases
and hence the kinetic energy decreases in going from B to A |
1 | 1604-1607 | (d) In moving a small negative charge from B to A work has to be
done by the external agency It is positive (e) Due to force of repulsion on the negative charge, velocity decreases
and hence the kinetic energy decreases in going from B to A EXAMPLE 2 |
1 | 1605-1608 | It is positive (e) Due to force of repulsion on the negative charge, velocity decreases
and hence the kinetic energy decreases in going from B to A EXAMPLE 2 3
Electric potential, equipotential surfaces:
http://video |
1 | 1606-1609 | (e) Due to force of repulsion on the negative charge, velocity decreases
and hence the kinetic energy decreases in going from B to A EXAMPLE 2 3
Electric potential, equipotential surfaces:
http://video mit |
1 | 1607-1610 | EXAMPLE 2 3
Electric potential, equipotential surfaces:
http://video mit edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field-
equipotential-sufaces-12584/
Rationalised 2023-24
Physics
54
FIGURE 2 |
1 | 1608-1611 | 3
Electric potential, equipotential surfaces:
http://video mit edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field-
equipotential-sufaces-12584/
Rationalised 2023-24
Physics
54
FIGURE 2 10 Equipotential surfaces for a uniform electric field |
1 | 1609-1612 | mit edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field-
equipotential-sufaces-12584/
Rationalised 2023-24
Physics
54
FIGURE 2 10 Equipotential surfaces for a uniform electric field 2 |
1 | 1610-1613 | edu/watch/4-electrostatic-potential-elctric-energy-ev-conservative-field-
equipotential-sufaces-12584/
Rationalised 2023-24
Physics
54
FIGURE 2 10 Equipotential surfaces for a uniform electric field 2 6 EQUIPOTENTIAL SURFACES
An equipotential surface is a surface with a constant value of potential
at all points on the surface |
1 | 1611-1614 | 10 Equipotential surfaces for a uniform electric field 2 6 EQUIPOTENTIAL SURFACES
An equipotential surface is a surface with a constant value of potential
at all points on the surface For a single charge q, the potential is given
by Eq |
1 | 1612-1615 | 2 6 EQUIPOTENTIAL SURFACES
An equipotential surface is a surface with a constant value of potential
at all points on the surface For a single charge q, the potential is given
by Eq (2 |
1 | 1613-1616 | 6 EQUIPOTENTIAL SURFACES
An equipotential surface is a surface with a constant value of potential
at all points on the surface For a single charge q, the potential is given
by Eq (2 8):
41
o
q
V
εr
=
π
This shows that V is a constant if r is constant |
1 | 1614-1617 | For a single charge q, the potential is given
by Eq (2 8):
41
o
q
V
εr
=
π
This shows that V is a constant if r is constant Thus, equipotential
surfaces of a single point charge are concentric spherical surfaces centred
at the charge |
1 | 1615-1618 | (2 8):
41
o
q
V
εr
=
π
This shows that V is a constant if r is constant Thus, equipotential
surfaces of a single point charge are concentric spherical surfaces centred
at the charge Now the electric field lines for a single charge q are radial lines starting
from or ending at the charge, depending on whether q is positive or negative |
1 | 1616-1619 | 8):
41
o
q
V
εr
=
π
This shows that V is a constant if r is constant Thus, equipotential
surfaces of a single point charge are concentric spherical surfaces centred
at the charge Now the electric field lines for a single charge q are radial lines starting
from or ending at the charge, depending on whether q is positive or negative Clearly, the electric field at every point is normal to the equipotential surface
passing through that point |
1 | 1617-1620 | Thus, equipotential
surfaces of a single point charge are concentric spherical surfaces centred
at the charge Now the electric field lines for a single charge q are radial lines starting
from or ending at the charge, depending on whether q is positive or negative Clearly, the electric field at every point is normal to the equipotential surface
passing through that point This is true in general: for any charge
configuration, equipotential surface through a point is normal to the
electric field at that point |
1 | 1618-1621 | Now the electric field lines for a single charge q are radial lines starting
from or ending at the charge, depending on whether q is positive or negative Clearly, the electric field at every point is normal to the equipotential surface
passing through that point This is true in general: for any charge
configuration, equipotential surface through a point is normal to the
electric field at that point The proof of this statement is simple |
1 | 1619-1622 | Clearly, the electric field at every point is normal to the equipotential surface
passing through that point This is true in general: for any charge
configuration, equipotential surface through a point is normal to the
electric field at that point The proof of this statement is simple If the field were not normal to the equipotential surface, it would
have non-zero component along the surface |
1 | 1620-1623 | This is true in general: for any charge
configuration, equipotential surface through a point is normal to the
electric field at that point The proof of this statement is simple If the field were not normal to the equipotential surface, it would
have non-zero component along the surface To move a unit test charge
against the direction of the component of the field, work would have to
be done |
1 | 1621-1624 | The proof of this statement is simple If the field were not normal to the equipotential surface, it would
have non-zero component along the surface To move a unit test charge
against the direction of the component of the field, work would have to
be done But this is in contradiction to the definition of an equipotential
surface: there is no potential difference between any two points on the
surface and no work is required to move a test charge on the surface |
1 | 1622-1625 | If the field were not normal to the equipotential surface, it would
have non-zero component along the surface To move a unit test charge
against the direction of the component of the field, work would have to
be done But this is in contradiction to the definition of an equipotential
surface: there is no potential difference between any two points on the
surface and no work is required to move a test charge on the surface The electric field must, therefore, be normal to the equipotential surface
at every point |
1 | 1623-1626 | To move a unit test charge
against the direction of the component of the field, work would have to
be done But this is in contradiction to the definition of an equipotential
surface: there is no potential difference between any two points on the
surface and no work is required to move a test charge on the surface The electric field must, therefore, be normal to the equipotential surface
at every point Equipotential surfaces offer an alternative visual picture
in addition to the picture of electric field lines around a charge
configuration |
1 | 1624-1627 | But this is in contradiction to the definition of an equipotential
surface: there is no potential difference between any two points on the
surface and no work is required to move a test charge on the surface The electric field must, therefore, be normal to the equipotential surface
at every point Equipotential surfaces offer an alternative visual picture
in addition to the picture of electric field lines around a charge
configuration FIGURE 2 |
1 | 1625-1628 | The electric field must, therefore, be normal to the equipotential surface
at every point Equipotential surfaces offer an alternative visual picture
in addition to the picture of electric field lines around a charge
configuration FIGURE 2 9 For a
single charge q
(a) equipotential
surfaces are
spherical surfaces
centred at the
charge, and
(b) electric field
lines are radial,
starting from the
charge if q > 0 |
1 | 1626-1629 | Equipotential surfaces offer an alternative visual picture
in addition to the picture of electric field lines around a charge
configuration FIGURE 2 9 For a
single charge q
(a) equipotential
surfaces are
spherical surfaces
centred at the
charge, and
(b) electric field
lines are radial,
starting from the
charge if q > 0 For a uniform electric field E, say, along the x-axis, the equipotential
surfaces are planes normal to the x-axis, i |
1 | 1627-1630 | FIGURE 2 9 For a
single charge q
(a) equipotential
surfaces are
spherical surfaces
centred at the
charge, and
(b) electric field
lines are radial,
starting from the
charge if q > 0 For a uniform electric field E, say, along the x-axis, the equipotential
surfaces are planes normal to the x-axis, i e |
1 | 1628-1631 | 9 For a
single charge q
(a) equipotential
surfaces are
spherical surfaces
centred at the
charge, and
(b) electric field
lines are radial,
starting from the
charge if q > 0 For a uniform electric field E, say, along the x-axis, the equipotential
surfaces are planes normal to the x-axis, i e , planes parallel to the y-z
plane (Fig |
1 | 1629-1632 | For a uniform electric field E, say, along the x-axis, the equipotential
surfaces are planes normal to the x-axis, i e , planes parallel to the y-z
plane (Fig 2 |
1 | 1630-1633 | e , planes parallel to the y-z
plane (Fig 2 10) |
1 | 1631-1634 | , planes parallel to the y-z
plane (Fig 2 10) Equipotential surfaces for (a) a dipole and (b) two
identical positive charges are shown in Fig |
1 | 1632-1635 | 2 10) Equipotential surfaces for (a) a dipole and (b) two
identical positive charges are shown in Fig 2 |
1 | 1633-1636 | 10) Equipotential surfaces for (a) a dipole and (b) two
identical positive charges are shown in Fig 2 11 |
1 | 1634-1637 | Equipotential surfaces for (a) a dipole and (b) two
identical positive charges are shown in Fig 2 11 FIGURE 2 |
1 | 1635-1638 | 2 11 FIGURE 2 11 Some equipotential surfaces for (a) a dipole,
(b) two identical positive charges |
1 | 1636-1639 | 11 FIGURE 2 11 Some equipotential surfaces for (a) a dipole,
(b) two identical positive charges Rationalised 2023-24
Electrostatic Potential
and Capacitance
55
2 |
1 | 1637-1640 | FIGURE 2 11 Some equipotential surfaces for (a) a dipole,
(b) two identical positive charges Rationalised 2023-24
Electrostatic Potential
and Capacitance
55
2 6 |
1 | 1638-1641 | 11 Some equipotential surfaces for (a) a dipole,
(b) two identical positive charges Rationalised 2023-24
Electrostatic Potential
and Capacitance
55
2 6 1 Relation between field and potential
Consider two closely spaced equipotential surfaces A and B (Fig |
1 | 1639-1642 | Rationalised 2023-24
Electrostatic Potential
and Capacitance
55
2 6 1 Relation between field and potential
Consider two closely spaced equipotential surfaces A and B (Fig 2 |
1 | 1640-1643 | 6 1 Relation between field and potential
Consider two closely spaced equipotential surfaces A and B (Fig 2 12)
with potential values V and V + d V, where d V is the change in V in the
direction of the electric field E |
1 | 1641-1644 | 1 Relation between field and potential
Consider two closely spaced equipotential surfaces A and B (Fig 2 12)
with potential values V and V + d V, where d V is the change in V in the
direction of the electric field E Let P be a point on the
surface B |
1 | 1642-1645 | 2 12)
with potential values V and V + d V, where d V is the change in V in the
direction of the electric field E Let P be a point on the
surface B d l is the perpendicular distance of the
surface A from P |
1 | 1643-1646 | 12)
with potential values V and V + d V, where d V is the change in V in the
direction of the electric field E Let P be a point on the
surface B d l is the perpendicular distance of the
surface A from P Imagine that a unit positive charge
is moved along this perpendicular from the surface B
to surface A against the electric field |
1 | 1644-1647 | Let P be a point on the
surface B d l is the perpendicular distance of the
surface A from P Imagine that a unit positive charge
is moved along this perpendicular from the surface B
to surface A against the electric field The work done
in this process is |E|d l |
1 | 1645-1648 | d l is the perpendicular distance of the
surface A from P Imagine that a unit positive charge
is moved along this perpendicular from the surface B
to surface A against the electric field The work done
in this process is |E|d l This work equals the potential difference
VA–VB |
1 | 1646-1649 | Imagine that a unit positive charge
is moved along this perpendicular from the surface B
to surface A against the electric field The work done
in this process is |E|d l This work equals the potential difference
VA–VB Thus,
|E|d l = V – (V + dV)= – dV
i |
1 | 1647-1650 | The work done
in this process is |E|d l This work equals the potential difference
VA–VB Thus,
|E|d l = V – (V + dV)= – dV
i e |
1 | 1648-1651 | This work equals the potential difference
VA–VB Thus,
|E|d l = V – (V + dV)= – dV
i e , |E|= − δ
δ
V
l
(2 |
1 | 1649-1652 | Thus,
|E|d l = V – (V + dV)= – dV
i e , |E|= − δ
δ
V
l
(2 20)
Since dV is negative, dV = – |dV| |
1 | 1650-1653 | e , |E|= − δ
δ
V
l
(2 20)
Since dV is negative, dV = – |dV| we can rewrite
Eq (2 |
1 | 1651-1654 | , |E|= − δ
δ
V
l
(2 20)
Since dV is negative, dV = – |dV| we can rewrite
Eq (2 20) as
E = −
= +
δδ
δδ
lV
lV
(2 |
1 | 1652-1655 | 20)
Since dV is negative, dV = – |dV| we can rewrite
Eq (2 20) as
E = −
= +
δδ
δδ
lV
lV
(2 21)
We thus arrive at two important conclusions concerning the relation
between electric field and potential:
(i)
Electric field is in the direction in which the potential decreases
steepest |
1 | 1653-1656 | we can rewrite
Eq (2 20) as
E = −
= +
δδ
δδ
lV
lV
(2 21)
We thus arrive at two important conclusions concerning the relation
between electric field and potential:
(i)
Electric field is in the direction in which the potential decreases
steepest (ii) Its magnitude is given by the change in the magnitude of potential
per unit displacement normal to the equipotential surface at the point |
1 | 1654-1657 | 20) as
E = −
= +
δδ
δδ
lV
lV
(2 21)
We thus arrive at two important conclusions concerning the relation
between electric field and potential:
(i)
Electric field is in the direction in which the potential decreases
steepest (ii) Its magnitude is given by the change in the magnitude of potential
per unit displacement normal to the equipotential surface at the point 2 |
1 | 1655-1658 | 21)
We thus arrive at two important conclusions concerning the relation
between electric field and potential:
(i)
Electric field is in the direction in which the potential decreases
steepest (ii) Its magnitude is given by the change in the magnitude of potential
per unit displacement normal to the equipotential surface at the point 2 7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Consider first the simple case of two charges q1and q2 with position vector
r1 and r2 relative to some origin |
1 | 1656-1659 | (ii) Its magnitude is given by the change in the magnitude of potential
per unit displacement normal to the equipotential surface at the point 2 7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Consider first the simple case of two charges q1and q2 with position vector
r1 and r2 relative to some origin Let us calculate the work done
(externally) in building up this configuration |
1 | 1657-1660 | 2 7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Consider first the simple case of two charges q1and q2 with position vector
r1 and r2 relative to some origin Let us calculate the work done
(externally) in building up this configuration This means that we consider
the charges q1 and q2 initially at infinity and determine the work done by
an external agency to bring the charges to the given locations |
1 | 1658-1661 | 7 POTENTIAL ENERGY OF A SYSTEM OF CHARGES
Consider first the simple case of two charges q1and q2 with position vector
r1 and r2 relative to some origin Let us calculate the work done
(externally) in building up this configuration This means that we consider
the charges q1 and q2 initially at infinity and determine the work done by
an external agency to bring the charges to the given locations Suppose,
first the charge q1 is brought from infinity to the point r1 |
1 | 1659-1662 | Let us calculate the work done
(externally) in building up this configuration This means that we consider
the charges q1 and q2 initially at infinity and determine the work done by
an external agency to bring the charges to the given locations Suppose,
first the charge q1 is brought from infinity to the point r1 There is no
external field against which work needs to be done, so work done in
bringing q1 from infinity to r1 is zero |
1 | 1660-1663 | This means that we consider
the charges q1 and q2 initially at infinity and determine the work done by
an external agency to bring the charges to the given locations Suppose,
first the charge q1 is brought from infinity to the point r1 There is no
external field against which work needs to be done, so work done in
bringing q1 from infinity to r1 is zero This charge produces a potential in
space given by
V
rq
1
0
1
1
=41
πε
P
where r1P is the distance of a point P in space from the location of q1 |
1 | 1661-1664 | Suppose,
first the charge q1 is brought from infinity to the point r1 There is no
external field against which work needs to be done, so work done in
bringing q1 from infinity to r1 is zero This charge produces a potential in
space given by
V
rq
1
0
1
1
=41
πε
P
where r1P is the distance of a point P in space from the location of q1 From the definition of potential, work done in bringing charge q2 from
infinity to the point r2 is q2 times the potential at r2 due to q1:
work done on q2 =
41
0
1
2
12
πε
q q
r
FIGURE 2 |
1 | 1662-1665 | There is no
external field against which work needs to be done, so work done in
bringing q1 from infinity to r1 is zero This charge produces a potential in
space given by
V
rq
1
0
1
1
=41
πε
P
where r1P is the distance of a point P in space from the location of q1 From the definition of potential, work done in bringing charge q2 from
infinity to the point r2 is q2 times the potential at r2 due to q1:
work done on q2 =
41
0
1
2
12
πε
q q
r
FIGURE 2 12 From the
potential to the field |
1 | 1663-1666 | This charge produces a potential in
space given by
V
rq
1
0
1
1
=41
πε
P
where r1P is the distance of a point P in space from the location of q1 From the definition of potential, work done in bringing charge q2 from
infinity to the point r2 is q2 times the potential at r2 due to q1:
work done on q2 =
41
0
1
2
12
πε
q q
r
FIGURE 2 12 From the
potential to the field Rationalised 2023-24
Physics
56
where r12 is the distance between points 1 and 2 |
1 | 1664-1667 | From the definition of potential, work done in bringing charge q2 from
infinity to the point r2 is q2 times the potential at r2 due to q1:
work done on q2 =
41
0
1
2
12
πε
q q
r
FIGURE 2 12 From the
potential to the field Rationalised 2023-24
Physics
56
where r12 is the distance between points 1 and 2 Since electrostatic force is conservative, this work gets
stored in the form of potential energy of the system |
1 | 1665-1668 | 12 From the
potential to the field Rationalised 2023-24
Physics
56
where r12 is the distance between points 1 and 2 Since electrostatic force is conservative, this work gets
stored in the form of potential energy of the system Thus,
the potential energy of a system of two charges q1 and q2 is
U
rq q
=
41
0
1
2
12
πε
(2 |
1 | 1666-1669 | Rationalised 2023-24
Physics
56
where r12 is the distance between points 1 and 2 Since electrostatic force is conservative, this work gets
stored in the form of potential energy of the system Thus,
the potential energy of a system of two charges q1 and q2 is
U
rq q
=
41
0
1
2
12
πε
(2 22)
Obviously, if q2 was brought first to its present location and
q1 brought later, the potential energy U would be the same |
1 | 1667-1670 | Since electrostatic force is conservative, this work gets
stored in the form of potential energy of the system Thus,
the potential energy of a system of two charges q1 and q2 is
U
rq q
=
41
0
1
2
12
πε
(2 22)
Obviously, if q2 was brought first to its present location and
q1 brought later, the potential energy U would be the same More generally, the potential energy expression,
Eq |
1 | 1668-1671 | Thus,
the potential energy of a system of two charges q1 and q2 is
U
rq q
=
41
0
1
2
12
πε
(2 22)
Obviously, if q2 was brought first to its present location and
q1 brought later, the potential energy U would be the same More generally, the potential energy expression,
Eq (2 |
1 | 1669-1672 | 22)
Obviously, if q2 was brought first to its present location and
q1 brought later, the potential energy U would be the same More generally, the potential energy expression,
Eq (2 22), is unaltered whatever way the charges are brought to the specified
locations, because of path-independence of work for electrostatic force |
1 | 1670-1673 | More generally, the potential energy expression,
Eq (2 22), is unaltered whatever way the charges are brought to the specified
locations, because of path-independence of work for electrostatic force Equation (2 |
1 | 1671-1674 | (2 22), is unaltered whatever way the charges are brought to the specified
locations, because of path-independence of work for electrostatic force Equation (2 22) is true for any sign of q1and q2 |
1 | 1672-1675 | 22), is unaltered whatever way the charges are brought to the specified
locations, because of path-independence of work for electrostatic force Equation (2 22) is true for any sign of q1and q2 If q1q2 > 0, potential
energy is positive |
1 | 1673-1676 | Equation (2 22) is true for any sign of q1and q2 If q1q2 > 0, potential
energy is positive This is as expected, since for like charges (q1q2 > 0),
electrostatic force is repulsive and a positive amount of work is needed to
be done against this force to bring the charges from infinity to a finite
distance apart |
1 | 1674-1677 | 22) is true for any sign of q1and q2 If q1q2 > 0, potential
energy is positive This is as expected, since for like charges (q1q2 > 0),
electrostatic force is repulsive and a positive amount of work is needed to
be done against this force to bring the charges from infinity to a finite
distance apart For unlike charges (q1 q2 < 0), the electrostatic force is
attractive |
1 | 1675-1678 | If q1q2 > 0, potential
energy is positive This is as expected, since for like charges (q1q2 > 0),
electrostatic force is repulsive and a positive amount of work is needed to
be done against this force to bring the charges from infinity to a finite
distance apart For unlike charges (q1 q2 < 0), the electrostatic force is
attractive In that case, a positive amount of work is needed against this
force to take the charges from the given location to infinity |
1 | 1676-1679 | This is as expected, since for like charges (q1q2 > 0),
electrostatic force is repulsive and a positive amount of work is needed to
be done against this force to bring the charges from infinity to a finite
distance apart For unlike charges (q1 q2 < 0), the electrostatic force is
attractive In that case, a positive amount of work is needed against this
force to take the charges from the given location to infinity In other words,
a negative amount of work is needed for the reverse path (from infinity to
the present locations), so the potential energy is negative |
1 | 1677-1680 | For unlike charges (q1 q2 < 0), the electrostatic force is
attractive In that case, a positive amount of work is needed against this
force to take the charges from the given location to infinity In other words,
a negative amount of work is needed for the reverse path (from infinity to
the present locations), so the potential energy is negative Equation (2 |
1 | 1678-1681 | In that case, a positive amount of work is needed against this
force to take the charges from the given location to infinity In other words,
a negative amount of work is needed for the reverse path (from infinity to
the present locations), so the potential energy is negative Equation (2 22) is easily generalised for a system of any number of
point charges |
1 | 1679-1682 | In other words,
a negative amount of work is needed for the reverse path (from infinity to
the present locations), so the potential energy is negative Equation (2 22) is easily generalised for a system of any number of
point charges Let us calculate the potential energy of a system of three
charges q1, q2 and q3 located at r1, r2, r3, respectively |
1 | 1680-1683 | Equation (2 22) is easily generalised for a system of any number of
point charges Let us calculate the potential energy of a system of three
charges q1, q2 and q3 located at r1, r2, r3, respectively To bring q1 first
from infinity to r1, no work is required |
1 | 1681-1684 | 22) is easily generalised for a system of any number of
point charges Let us calculate the potential energy of a system of three
charges q1, q2 and q3 located at r1, r2, r3, respectively To bring q1 first
from infinity to r1, no work is required Next we bring q2 from infinity to
r2 |
1 | 1682-1685 | Let us calculate the potential energy of a system of three
charges q1, q2 and q3 located at r1, r2, r3, respectively To bring q1 first
from infinity to r1, no work is required Next we bring q2 from infinity to
r2 As before, work done in this step is
1
2
2
1
2
0
12
1
(
)
4
q q
q V
r
ε
=
π
r
(2 |
1 | 1683-1686 | To bring q1 first
from infinity to r1, no work is required Next we bring q2 from infinity to
r2 As before, work done in this step is
1
2
2
1
2
0
12
1
(
)
4
q q
q V
r
ε
=
π
r
(2 23)
The charges q1 and q2 produce a potential, which at any point P is
given by
V
rq
rq
1 2
0
1
1
2
2
41
,
=
+
πε
P
P
(2 |
1 | 1684-1687 | Next we bring q2 from infinity to
r2 As before, work done in this step is
1
2
2
1
2
0
12
1
(
)
4
q q
q V
r
ε
=
π
r
(2 23)
The charges q1 and q2 produce a potential, which at any point P is
given by
V
rq
rq
1 2
0
1
1
2
2
41
,
=
+
πε
P
P
(2 24)
Work done next in bringing q3 from infinity to the point r3 is q3 times
V1, 2 at r3
q V
rq q
q q
r
3
1 2
3
0
1
3
13
2
3
23
41
, (
r)
=
+
πε
(2 |
1 | 1685-1688 | As before, work done in this step is
1
2
2
1
2
0
12
1
(
)
4
q q
q V
r
ε
=
π
r
(2 23)
The charges q1 and q2 produce a potential, which at any point P is
given by
V
rq
rq
1 2
0
1
1
2
2
41
,
=
+
πε
P
P
(2 24)
Work done next in bringing q3 from infinity to the point r3 is q3 times
V1, 2 at r3
q V
rq q
q q
r
3
1 2
3
0
1
3
13
2
3
23
41
, (
r)
=
+
πε
(2 25)
The total work done in assembling the charges
at the given locations is obtained by adding the work
done in different steps [Eq |
1 | 1686-1689 | 23)
The charges q1 and q2 produce a potential, which at any point P is
given by
V
rq
rq
1 2
0
1
1
2
2
41
,
=
+
πε
P
P
(2 24)
Work done next in bringing q3 from infinity to the point r3 is q3 times
V1, 2 at r3
q V
rq q
q q
r
3
1 2
3
0
1
3
13
2
3
23
41
, (
r)
=
+
πε
(2 25)
The total work done in assembling the charges
at the given locations is obtained by adding the work
done in different steps [Eq (2 |
1 | 1687-1690 | 24)
Work done next in bringing q3 from infinity to the point r3 is q3 times
V1, 2 at r3
q V
rq q
q q
r
3
1 2
3
0
1
3
13
2
3
23
41
, (
r)
=
+
πε
(2 25)
The total work done in assembling the charges
at the given locations is obtained by adding the work
done in different steps [Eq (2 23) and Eq |
1 | 1688-1691 | 25)
The total work done in assembling the charges
at the given locations is obtained by adding the work
done in different steps [Eq (2 23) and Eq (2 |
1 | 1689-1692 | (2 23) and Eq (2 25)],
U
rq q
rq q
q q
r
=
+
+
41
0
1
2
12
1
3
13
2
3
23
πε
(2 |
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