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1 | 990-993 | 25) (a) What is the net outward flux through each
flat face (b) What is the flux through the side of the cylinder (c) What is the net outward flux through the cylinder |
1 | 991-994 | (a) What is the net outward flux through each
flat face (b) What is the flux through the side of the cylinder (c) What is the net outward flux through the cylinder (d) What is the
net charge inside the cylinder |
1 | 992-995 | (b) What is the flux through the side of the cylinder (c) What is the net outward flux through the cylinder (d) What is the
net charge inside the cylinder Solution
(a)
We can see from the figure that on the left face E and DS are
parallel |
1 | 993-996 | (c) What is the net outward flux through the cylinder (d) What is the
net charge inside the cylinder Solution
(a)
We can see from the figure that on the left face E and DS are
parallel Therefore, the outward flux is
fL= E |
1 | 994-997 | (d) What is the
net charge inside the cylinder Solution
(a)
We can see from the figure that on the left face E and DS are
parallel Therefore, the outward flux is
fL= E DS = – 200 ˆ ∆
i
S
i
= + 200 DS, since ˆ ∆
i
iS
= – DS
= + 200 × p (0 |
1 | 995-998 | Solution
(a)
We can see from the figure that on the left face E and DS are
parallel Therefore, the outward flux is
fL= E DS = – 200 ˆ ∆
i
S
i
= + 200 DS, since ˆ ∆
i
iS
= – DS
= + 200 × p (0 05)2 = + 1 |
1 | 996-999 | Therefore, the outward flux is
fL= E DS = – 200 ˆ ∆
i
S
i
= + 200 DS, since ˆ ∆
i
iS
= – DS
= + 200 × p (0 05)2 = + 1 57 N m2 C–1
On the right face, E and DS are parallel and therefore
fR = E |
1 | 997-1000 | DS = – 200 ˆ ∆
i
S
i
= + 200 DS, since ˆ ∆
i
iS
= – DS
= + 200 × p (0 05)2 = + 1 57 N m2 C–1
On the right face, E and DS are parallel and therefore
fR = E DS = + 1 |
1 | 998-1001 | 05)2 = + 1 57 N m2 C–1
On the right face, E and DS are parallel and therefore
fR = E DS = + 1 57 N m2 C–1 |
1 | 999-1002 | 57 N m2 C–1
On the right face, E and DS are parallel and therefore
fR = E DS = + 1 57 N m2 C–1 (b)
For any point on the side of the cylinder E is perpendicular to
DS and hence E |
1 | 1000-1003 | DS = + 1 57 N m2 C–1 (b)
For any point on the side of the cylinder E is perpendicular to
DS and hence E DS = 0 |
1 | 1001-1004 | 57 N m2 C–1 (b)
For any point on the side of the cylinder E is perpendicular to
DS and hence E DS = 0 Therefore, the flux out of the side of the
cylinder is zero |
1 | 1002-1005 | (b)
For any point on the side of the cylinder E is perpendicular to
DS and hence E DS = 0 Therefore, the flux out of the side of the
cylinder is zero (c)
Net outward flux through the cylinder
f = 1 |
1 | 1003-1006 | DS = 0 Therefore, the flux out of the side of the
cylinder is zero (c)
Net outward flux through the cylinder
f = 1 57 + 1 |
1 | 1004-1007 | Therefore, the flux out of the side of the
cylinder is zero (c)
Net outward flux through the cylinder
f = 1 57 + 1 57 + 0 = 3 |
1 | 1005-1008 | (c)
Net outward flux through the cylinder
f = 1 57 + 1 57 + 0 = 3 14 N m2 C–1
FIGURE 1 |
1 | 1006-1009 | 57 + 1 57 + 0 = 3 14 N m2 C–1
FIGURE 1 25
(d) The net charge within the cylinder can be found by using Gauss’s
law which gives
q = e0f
= 3 |
1 | 1007-1010 | 57 + 0 = 3 14 N m2 C–1
FIGURE 1 25
(d) The net charge within the cylinder can be found by using Gauss’s
law which gives
q = e0f
= 3 14 × 8 |
1 | 1008-1011 | 14 N m2 C–1
FIGURE 1 25
(d) The net charge within the cylinder can be found by using Gauss’s
law which gives
q = e0f
= 3 14 × 8 854 × 10–12 C
= 2 |
1 | 1009-1012 | 25
(d) The net charge within the cylinder can be found by using Gauss’s
law which gives
q = e0f
= 3 14 × 8 854 × 10–12 C
= 2 78 × 10–11 C
Rationalised 2023-24
Electric Charges
and Fields
33
1 |
1 | 1010-1013 | 14 × 8 854 × 10–12 C
= 2 78 × 10–11 C
Rationalised 2023-24
Electric Charges
and Fields
33
1 14 APPLICATIONS OF GAUSS’S LAW
The electric field due to a general charge distribution is, as seen above,
given by Eq |
1 | 1011-1014 | 854 × 10–12 C
= 2 78 × 10–11 C
Rationalised 2023-24
Electric Charges
and Fields
33
1 14 APPLICATIONS OF GAUSS’S LAW
The electric field due to a general charge distribution is, as seen above,
given by Eq (1 |
1 | 1012-1015 | 78 × 10–11 C
Rationalised 2023-24
Electric Charges
and Fields
33
1 14 APPLICATIONS OF GAUSS’S LAW
The electric field due to a general charge distribution is, as seen above,
given by Eq (1 27) |
1 | 1013-1016 | 14 APPLICATIONS OF GAUSS’S LAW
The electric field due to a general charge distribution is, as seen above,
given by Eq (1 27) In practice, except for some special cases, the
summation (or integration) involved in this equation cannot be carried
out to give electric field at every point in
space |
1 | 1014-1017 | (1 27) In practice, except for some special cases, the
summation (or integration) involved in this equation cannot be carried
out to give electric field at every point in
space For some symmetric charge
configurations, however, it is possible to
obtain the electric field in a simple way using
the Gauss’s law |
1 | 1015-1018 | 27) In practice, except for some special cases, the
summation (or integration) involved in this equation cannot be carried
out to give electric field at every point in
space For some symmetric charge
configurations, however, it is possible to
obtain the electric field in a simple way using
the Gauss’s law This is best understood by
some examples |
1 | 1016-1019 | In practice, except for some special cases, the
summation (or integration) involved in this equation cannot be carried
out to give electric field at every point in
space For some symmetric charge
configurations, however, it is possible to
obtain the electric field in a simple way using
the Gauss’s law This is best understood by
some examples 1 |
1 | 1017-1020 | For some symmetric charge
configurations, however, it is possible to
obtain the electric field in a simple way using
the Gauss’s law This is best understood by
some examples 1 14 |
1 | 1018-1021 | This is best understood by
some examples 1 14 1
Field due to an infinitely
long straight uniformly
charged wire
Consider an infinitely long thin straight wire
with uniform linear charge density l |
1 | 1019-1022 | 1 14 1
Field due to an infinitely
long straight uniformly
charged wire
Consider an infinitely long thin straight wire
with uniform linear charge density l The wire
is obviously an axis of symmetry |
1 | 1020-1023 | 14 1
Field due to an infinitely
long straight uniformly
charged wire
Consider an infinitely long thin straight wire
with uniform linear charge density l The wire
is obviously an axis of symmetry Suppose we
take the radial vector from O to P and rotate it
around the wire |
1 | 1021-1024 | 1
Field due to an infinitely
long straight uniformly
charged wire
Consider an infinitely long thin straight wire
with uniform linear charge density l The wire
is obviously an axis of symmetry Suppose we
take the radial vector from O to P and rotate it
around the wire The points P, P¢, P¢¢ so
obtained are completely equivalent with
respect to the charged wire |
1 | 1022-1025 | The wire
is obviously an axis of symmetry Suppose we
take the radial vector from O to P and rotate it
around the wire The points P, P¢, P¢¢ so
obtained are completely equivalent with
respect to the charged wire This implies that
the electric field must have the same magnitude
at these points |
1 | 1023-1026 | Suppose we
take the radial vector from O to P and rotate it
around the wire The points P, P¢, P¢¢ so
obtained are completely equivalent with
respect to the charged wire This implies that
the electric field must have the same magnitude
at these points The direction of electric field at
every point must be radial (outward if l > 0,
inward if l < 0) |
1 | 1024-1027 | The points P, P¢, P¢¢ so
obtained are completely equivalent with
respect to the charged wire This implies that
the electric field must have the same magnitude
at these points The direction of electric field at
every point must be radial (outward if l > 0,
inward if l < 0) This is clear from Fig |
1 | 1025-1028 | This implies that
the electric field must have the same magnitude
at these points The direction of electric field at
every point must be radial (outward if l > 0,
inward if l < 0) This is clear from Fig 1 |
1 | 1026-1029 | The direction of electric field at
every point must be radial (outward if l > 0,
inward if l < 0) This is clear from Fig 1 26 |
1 | 1027-1030 | This is clear from Fig 1 26 Consider a pair of line elements P1 and P2
of the wire, as shown |
1 | 1028-1031 | 1 26 Consider a pair of line elements P1 and P2
of the wire, as shown The electric fields
produced by the two elements of the pair when
summed give a resultant electric field which
is radial (the components normal to the radial
vector cancel) |
1 | 1029-1032 | 26 Consider a pair of line elements P1 and P2
of the wire, as shown The electric fields
produced by the two elements of the pair when
summed give a resultant electric field which
is radial (the components normal to the radial
vector cancel) This is true for any such pair
and hence the total field at any point P is
radial |
1 | 1030-1033 | Consider a pair of line elements P1 and P2
of the wire, as shown The electric fields
produced by the two elements of the pair when
summed give a resultant electric field which
is radial (the components normal to the radial
vector cancel) This is true for any such pair
and hence the total field at any point P is
radial Finally, since the wire is infinite,
electric field does not depend on the position
of P along the length of the wire |
1 | 1031-1034 | The electric fields
produced by the two elements of the pair when
summed give a resultant electric field which
is radial (the components normal to the radial
vector cancel) This is true for any such pair
and hence the total field at any point P is
radial Finally, since the wire is infinite,
electric field does not depend on the position
of P along the length of the wire In short, the
electric field is everywhere radial in the plane
cutting the wire normally, and its magnitude
depends only on the radial distance r |
1 | 1032-1035 | This is true for any such pair
and hence the total field at any point P is
radial Finally, since the wire is infinite,
electric field does not depend on the position
of P along the length of the wire In short, the
electric field is everywhere radial in the plane
cutting the wire normally, and its magnitude
depends only on the radial distance r To calculate the field, imagine a cylindrical
Gaussian surface, as shown in the Fig |
1 | 1033-1036 | Finally, since the wire is infinite,
electric field does not depend on the position
of P along the length of the wire In short, the
electric field is everywhere radial in the plane
cutting the wire normally, and its magnitude
depends only on the radial distance r To calculate the field, imagine a cylindrical
Gaussian surface, as shown in the Fig 1 |
1 | 1034-1037 | In short, the
electric field is everywhere radial in the plane
cutting the wire normally, and its magnitude
depends only on the radial distance r To calculate the field, imagine a cylindrical
Gaussian surface, as shown in the Fig 1 26(b) |
1 | 1035-1038 | To calculate the field, imagine a cylindrical
Gaussian surface, as shown in the Fig 1 26(b) Since the field is everywhere radial, flux
through the two ends of the cylindrical
Gaussian surface is zero |
1 | 1036-1039 | 1 26(b) Since the field is everywhere radial, flux
through the two ends of the cylindrical
Gaussian surface is zero At the cylindrical
part of the surface, E is normal to the surface
at every point, and its magnitude is constant,
since it depends only on r |
1 | 1037-1040 | 26(b) Since the field is everywhere radial, flux
through the two ends of the cylindrical
Gaussian surface is zero At the cylindrical
part of the surface, E is normal to the surface
at every point, and its magnitude is constant,
since it depends only on r The surface area
of the curved part is 2prl, where l is the length
of the cylinder |
1 | 1038-1041 | Since the field is everywhere radial, flux
through the two ends of the cylindrical
Gaussian surface is zero At the cylindrical
part of the surface, E is normal to the surface
at every point, and its magnitude is constant,
since it depends only on r The surface area
of the curved part is 2prl, where l is the length
of the cylinder FIGURE 1 |
1 | 1039-1042 | At the cylindrical
part of the surface, E is normal to the surface
at every point, and its magnitude is constant,
since it depends only on r The surface area
of the curved part is 2prl, where l is the length
of the cylinder FIGURE 1 26 (a) Electric field due to an
infinitely long thin straight wire is radial,
(b) The Gaussian surface for a long thin
wire of uniform linear charge density |
1 | 1040-1043 | The surface area
of the curved part is 2prl, where l is the length
of the cylinder FIGURE 1 26 (a) Electric field due to an
infinitely long thin straight wire is radial,
(b) The Gaussian surface for a long thin
wire of uniform linear charge density Rationalised 2023-24
34
Physics
Flux through the Gaussian surface
=
flux through the curved cylindrical part of the surface
=
E × 2prl
The surface includes charge equal to l l |
1 | 1041-1044 | FIGURE 1 26 (a) Electric field due to an
infinitely long thin straight wire is radial,
(b) The Gaussian surface for a long thin
wire of uniform linear charge density Rationalised 2023-24
34
Physics
Flux through the Gaussian surface
=
flux through the curved cylindrical part of the surface
=
E × 2prl
The surface includes charge equal to l l Gauss’s law then gives
E × 2prl = ll/e0
i |
1 | 1042-1045 | 26 (a) Electric field due to an
infinitely long thin straight wire is radial,
(b) The Gaussian surface for a long thin
wire of uniform linear charge density Rationalised 2023-24
34
Physics
Flux through the Gaussian surface
=
flux through the curved cylindrical part of the surface
=
E × 2prl
The surface includes charge equal to l l Gauss’s law then gives
E × 2prl = ll/e0
i e |
1 | 1043-1046 | Rationalised 2023-24
34
Physics
Flux through the Gaussian surface
=
flux through the curved cylindrical part of the surface
=
E × 2prl
The surface includes charge equal to l l Gauss’s law then gives
E × 2prl = ll/e0
i e ,E =
0
2
r
ελ
π
Vectorially, E at any point is given by
0
ˆ
2
r
ελ
=
π
E
n
(1 |
1 | 1044-1047 | Gauss’s law then gives
E × 2prl = ll/e0
i e ,E =
0
2
r
ελ
π
Vectorially, E at any point is given by
0
ˆ
2
r
ελ
=
π
E
n
(1 32)
where ˆn is the radial unit vector in the plane normal to the wire passing
through the point |
1 | 1045-1048 | e ,E =
0
2
r
ελ
π
Vectorially, E at any point is given by
0
ˆ
2
r
ελ
=
π
E
n
(1 32)
where ˆn is the radial unit vector in the plane normal to the wire passing
through the point E is directed outward if l is positive and inward if l is
negative |
1 | 1046-1049 | ,E =
0
2
r
ελ
π
Vectorially, E at any point is given by
0
ˆ
2
r
ελ
=
π
E
n
(1 32)
where ˆn is the radial unit vector in the plane normal to the wire passing
through the point E is directed outward if l is positive and inward if l is
negative Note that when we write a vector A as a scalar multiplied by a unit
vector, i |
1 | 1047-1050 | 32)
where ˆn is the radial unit vector in the plane normal to the wire passing
through the point E is directed outward if l is positive and inward if l is
negative Note that when we write a vector A as a scalar multiplied by a unit
vector, i e |
1 | 1048-1051 | E is directed outward if l is positive and inward if l is
negative Note that when we write a vector A as a scalar multiplied by a unit
vector, i e , as A = A ˆa , the scalar A is an algebraic number |
1 | 1049-1052 | Note that when we write a vector A as a scalar multiplied by a unit
vector, i e , as A = A ˆa , the scalar A is an algebraic number It can be
negative or positive |
1 | 1050-1053 | e , as A = A ˆa , the scalar A is an algebraic number It can be
negative or positive The direction of A will be the same as that of the unit
vector ˆa if A > 0 and opposite to ˆa if A < 0 |
1 | 1051-1054 | , as A = A ˆa , the scalar A is an algebraic number It can be
negative or positive The direction of A will be the same as that of the unit
vector ˆa if A > 0 and opposite to ˆa if A < 0 When we want to restrict to
non-negative values, we use the symbol A and call it the modulus of A |
1 | 1052-1055 | It can be
negative or positive The direction of A will be the same as that of the unit
vector ˆa if A > 0 and opposite to ˆa if A < 0 When we want to restrict to
non-negative values, we use the symbol A and call it the modulus of A Thus,
A≥0 |
1 | 1053-1056 | The direction of A will be the same as that of the unit
vector ˆa if A > 0 and opposite to ˆa if A < 0 When we want to restrict to
non-negative values, we use the symbol A and call it the modulus of A Thus,
A≥0 Also note that though only the charge enclosed by the surface (ll)
was included above, the electric field E is due to the charge on the entire
wire |
1 | 1054-1057 | When we want to restrict to
non-negative values, we use the symbol A and call it the modulus of A Thus,
A≥0 Also note that though only the charge enclosed by the surface (ll)
was included above, the electric field E is due to the charge on the entire
wire Further, the assumption that the wire is infinitely long is crucial |
1 | 1055-1058 | Thus,
A≥0 Also note that though only the charge enclosed by the surface (ll)
was included above, the electric field E is due to the charge on the entire
wire Further, the assumption that the wire is infinitely long is crucial Without this assumption, we cannot take E to be normal to the curved
part of the cylindrical Gaussian surface |
1 | 1056-1059 | Also note that though only the charge enclosed by the surface (ll)
was included above, the electric field E is due to the charge on the entire
wire Further, the assumption that the wire is infinitely long is crucial Without this assumption, we cannot take E to be normal to the curved
part of the cylindrical Gaussian surface However, Eq |
1 | 1057-1060 | Further, the assumption that the wire is infinitely long is crucial Without this assumption, we cannot take E to be normal to the curved
part of the cylindrical Gaussian surface However, Eq (1 |
1 | 1058-1061 | Without this assumption, we cannot take E to be normal to the curved
part of the cylindrical Gaussian surface However, Eq (1 32) is
approximately true for electric field around the central portions of a long
wire, where the end effects may be ignored |
1 | 1059-1062 | However, Eq (1 32) is
approximately true for electric field around the central portions of a long
wire, where the end effects may be ignored 1 |
1 | 1060-1063 | (1 32) is
approximately true for electric field around the central portions of a long
wire, where the end effects may be ignored 1 14 |
1 | 1061-1064 | 32) is
approximately true for electric field around the central portions of a long
wire, where the end effects may be ignored 1 14 2 Field due to a uniformly charged infinite plane sheet
Let s be the uniform surface charge density of an infinite plane sheet
(Fig |
1 | 1062-1065 | 1 14 2 Field due to a uniformly charged infinite plane sheet
Let s be the uniform surface charge density of an infinite plane sheet
(Fig 1 |
1 | 1063-1066 | 14 2 Field due to a uniformly charged infinite plane sheet
Let s be the uniform surface charge density of an infinite plane sheet
(Fig 1 27) |
1 | 1064-1067 | 2 Field due to a uniformly charged infinite plane sheet
Let s be the uniform surface charge density of an infinite plane sheet
(Fig 1 27) We take the x-axis normal to the given plane |
1 | 1065-1068 | 1 27) We take the x-axis normal to the given plane By symmetry,
the electric field will not depend on y and z coordinates and its direction
at every point must be parallel to the x-direction |
1 | 1066-1069 | 27) We take the x-axis normal to the given plane By symmetry,
the electric field will not depend on y and z coordinates and its direction
at every point must be parallel to the x-direction We can take the Gaussian surface to be a
rectangular parallelepiped of cross-sectional area
A, as shown |
1 | 1067-1070 | We take the x-axis normal to the given plane By symmetry,
the electric field will not depend on y and z coordinates and its direction
at every point must be parallel to the x-direction We can take the Gaussian surface to be a
rectangular parallelepiped of cross-sectional area
A, as shown (A cylindrical surface will also do |
1 | 1068-1071 | By symmetry,
the electric field will not depend on y and z coordinates and its direction
at every point must be parallel to the x-direction We can take the Gaussian surface to be a
rectangular parallelepiped of cross-sectional area
A, as shown (A cylindrical surface will also do ) As
seen from the figure, only the two faces 1 and 2 will
contribute to the flux; electric field lines are parallel
to the other faces and they, therefore, do not
contribute to the total flux |
1 | 1069-1072 | We can take the Gaussian surface to be a
rectangular parallelepiped of cross-sectional area
A, as shown (A cylindrical surface will also do ) As
seen from the figure, only the two faces 1 and 2 will
contribute to the flux; electric field lines are parallel
to the other faces and they, therefore, do not
contribute to the total flux The unit vector normal to surface 1 is in –x
direction while the unit vector normal to surface 2
is in the +x direction |
1 | 1070-1073 | (A cylindrical surface will also do ) As
seen from the figure, only the two faces 1 and 2 will
contribute to the flux; electric field lines are parallel
to the other faces and they, therefore, do not
contribute to the total flux The unit vector normal to surface 1 is in –x
direction while the unit vector normal to surface 2
is in the +x direction Therefore, flux E |
1 | 1071-1074 | ) As
seen from the figure, only the two faces 1 and 2 will
contribute to the flux; electric field lines are parallel
to the other faces and they, therefore, do not
contribute to the total flux The unit vector normal to surface 1 is in –x
direction while the unit vector normal to surface 2
is in the +x direction Therefore, flux E DS through
both the surfaces are equal and add up |
1 | 1072-1075 | The unit vector normal to surface 1 is in –x
direction while the unit vector normal to surface 2
is in the +x direction Therefore, flux E DS through
both the surfaces are equal and add up Therefore
the net flux through the Gaussian surface is 2 EA |
1 | 1073-1076 | Therefore, flux E DS through
both the surfaces are equal and add up Therefore
the net flux through the Gaussian surface is 2 EA The charge enclosed by the closed surface is sA |
1 | 1074-1077 | DS through
both the surfaces are equal and add up Therefore
the net flux through the Gaussian surface is 2 EA The charge enclosed by the closed surface is sA Therefore by Gauss’s law,
FIGURE 1 |
1 | 1075-1078 | Therefore
the net flux through the Gaussian surface is 2 EA The charge enclosed by the closed surface is sA Therefore by Gauss’s law,
FIGURE 1 27 Gaussian surface for a
uniformly charged infinite plane sheet |
1 | 1076-1079 | The charge enclosed by the closed surface is sA Therefore by Gauss’s law,
FIGURE 1 27 Gaussian surface for a
uniformly charged infinite plane sheet Rationalised 2023-24
Electric Charges
and Fields
35
2 EA = sA/e0
or, E = s/2e0
Vectorically,
0
ˆ
2
εσ
E=
n
(1 |
1 | 1077-1080 | Therefore by Gauss’s law,
FIGURE 1 27 Gaussian surface for a
uniformly charged infinite plane sheet Rationalised 2023-24
Electric Charges
and Fields
35
2 EA = sA/e0
or, E = s/2e0
Vectorically,
0
ˆ
2
εσ
E=
n
(1 33)
where ˆn is a unit vector normal to the plane and going away from it |
1 | 1078-1081 | 27 Gaussian surface for a
uniformly charged infinite plane sheet Rationalised 2023-24
Electric Charges
and Fields
35
2 EA = sA/e0
or, E = s/2e0
Vectorically,
0
ˆ
2
εσ
E=
n
(1 33)
where ˆn is a unit vector normal to the plane and going away from it E is directed away from the plate if s is positive and toward the plate
if s is negative |
1 | 1079-1082 | Rationalised 2023-24
Electric Charges
and Fields
35
2 EA = sA/e0
or, E = s/2e0
Vectorically,
0
ˆ
2
εσ
E=
n
(1 33)
where ˆn is a unit vector normal to the plane and going away from it E is directed away from the plate if s is positive and toward the plate
if s is negative Note that the above application of the Gauss’ law has
brought out an additional fact: E is independent of x also |
1 | 1080-1083 | 33)
where ˆn is a unit vector normal to the plane and going away from it E is directed away from the plate if s is positive and toward the plate
if s is negative Note that the above application of the Gauss’ law has
brought out an additional fact: E is independent of x also For a finite large planar sheet, Eq |
1 | 1081-1084 | E is directed away from the plate if s is positive and toward the plate
if s is negative Note that the above application of the Gauss’ law has
brought out an additional fact: E is independent of x also For a finite large planar sheet, Eq (1 |
1 | 1082-1085 | Note that the above application of the Gauss’ law has
brought out an additional fact: E is independent of x also For a finite large planar sheet, Eq (1 33) is approximately true in the
middle regions of the planar sheet, away from the ends |
1 | 1083-1086 | For a finite large planar sheet, Eq (1 33) is approximately true in the
middle regions of the planar sheet, away from the ends 1 |
1 | 1084-1087 | (1 33) is approximately true in the
middle regions of the planar sheet, away from the ends 1 14 |
1 | 1085-1088 | 33) is approximately true in the
middle regions of the planar sheet, away from the ends 1 14 3 Field due to a uniformly charged thin spherical shell
Let s be the uniform surface charge density of a thin spherical shell of
radius R (Fig |
1 | 1086-1089 | 1 14 3 Field due to a uniformly charged thin spherical shell
Let s be the uniform surface charge density of a thin spherical shell of
radius R (Fig 1 |
1 | 1087-1090 | 14 3 Field due to a uniformly charged thin spherical shell
Let s be the uniform surface charge density of a thin spherical shell of
radius R (Fig 1 28) |
1 | 1088-1091 | 3 Field due to a uniformly charged thin spherical shell
Let s be the uniform surface charge density of a thin spherical shell of
radius R (Fig 1 28) The situation has obvious spherical symmetry |
1 | 1089-1092 | 1 28) The situation has obvious spherical symmetry The
field at any point P, outside or inside, can depend only on r (the radial
distance from the centre of the shell to the point) and must be radial (i |
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