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1 | 1090-1093 | 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 e |
1 | 1091-1094 | 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 e ,
along the radius vector) |
1 | 1092-1095 | 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 e ,
along the radius vector) (i) Field outside the shell: Consider a point P outside the
shell with radius vector r |
1 | 1093-1096 | e ,
along the radius vector) (i) Field outside the shell: Consider a point P outside the
shell with radius vector r To calculate E at P, we take the
Gaussian surface to be a sphere of radius r and with centre
O, passing through P |
1 | 1094-1097 | ,
along the radius vector) (i) Field outside the shell: Consider a point P outside the
shell with radius vector r To calculate E at P, we take the
Gaussian surface to be a sphere of radius r and with centre
O, passing through P All points on this sphere are equivalent
relative to the given charged configuration |
1 | 1095-1098 | (i) Field outside the shell: Consider a point P outside the
shell with radius vector r To calculate E at P, we take the
Gaussian surface to be a sphere of radius r and with centre
O, passing through P All points on this sphere are equivalent
relative to the given charged configuration (That is what we
mean by spherical symmetry |
1 | 1096-1099 | To calculate E at P, we take the
Gaussian surface to be a sphere of radius r and with centre
O, passing through P All points on this sphere are equivalent
relative to the given charged configuration (That is what we
mean by spherical symmetry ) The electric field at each point
of the Gaussian surface, therefore, has the same magnitude
E and is along the radius vector at each point |
1 | 1097-1100 | All points on this sphere are equivalent
relative to the given charged configuration (That is what we
mean by spherical symmetry ) The electric field at each point
of the Gaussian surface, therefore, has the same magnitude
E and is along the radius vector at each point Thus, E and
DS at every point are parallel and the flux through each
element is E DS |
1 | 1098-1101 | (That is what we
mean by spherical symmetry ) The electric field at each point
of the Gaussian surface, therefore, has the same magnitude
E and is along the radius vector at each point Thus, E and
DS at every point are parallel and the flux through each
element is E DS Summing over all DS, the flux through the
Gaussian surface is E × 4 p r 2 |
1 | 1099-1102 | ) The electric field at each point
of the Gaussian surface, therefore, has the same magnitude
E and is along the radius vector at each point Thus, E and
DS at every point are parallel and the flux through each
element is E DS Summing over all DS, the flux through the
Gaussian surface is E × 4 p r 2 The charge enclosed is
s × 4 p R 2 |
1 | 1100-1103 | Thus, E and
DS at every point are parallel and the flux through each
element is E DS Summing over all DS, the flux through the
Gaussian surface is E × 4 p r 2 The charge enclosed is
s × 4 p R 2 By Gauss’s law
E × 4 p r 2 =
2
0
4
R
εσ
π
Or,
2
2
2
0
0
4
R
q
E
r
r
εσ
ε
=
=
π
where q = 4 p R2 s is the total charge on the spherical shell |
1 | 1101-1104 | Summing over all DS, the flux through the
Gaussian surface is E × 4 p r 2 The charge enclosed is
s × 4 p R 2 By Gauss’s law
E × 4 p r 2 =
2
0
4
R
εσ
π
Or,
2
2
2
0
0
4
R
q
E
r
r
εσ
ε
=
=
π
where q = 4 p R2 s is the total charge on the spherical shell Vectorially,
2
0
ˆ
4
q
εr
=
π
E
r
(1 |
1 | 1102-1105 | The charge enclosed is
s × 4 p R 2 By Gauss’s law
E × 4 p r 2 =
2
0
4
R
εσ
π
Or,
2
2
2
0
0
4
R
q
E
r
r
εσ
ε
=
=
π
where q = 4 p R2 s is the total charge on the spherical shell Vectorially,
2
0
ˆ
4
q
εr
=
π
E
r
(1 34)
The electric field is directed outward if q > 0 and inward if
q < 0 |
1 | 1103-1106 | By Gauss’s law
E × 4 p r 2 =
2
0
4
R
εσ
π
Or,
2
2
2
0
0
4
R
q
E
r
r
εσ
ε
=
=
π
where q = 4 p R2 s is the total charge on the spherical shell Vectorially,
2
0
ˆ
4
q
εr
=
π
E
r
(1 34)
The electric field is directed outward if q > 0 and inward if
q < 0 This, however, is exactly the field produced by a charge
q placed at the centre O |
1 | 1104-1107 | Vectorially,
2
0
ˆ
4
q
εr
=
π
E
r
(1 34)
The electric field is directed outward if q > 0 and inward if
q < 0 This, however, is exactly the field produced by a charge
q placed at the centre O Thus for points outside the shell, the field due
to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre |
1 | 1105-1108 | 34)
The electric field is directed outward if q > 0 and inward if
q < 0 This, however, is exactly the field produced by a charge
q placed at the centre O Thus for points outside the shell, the field due
to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre (ii) Field inside the shell: In Fig |
1 | 1106-1109 | This, however, is exactly the field produced by a charge
q placed at the centre O Thus for points outside the shell, the field due
to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre (ii) Field inside the shell: In Fig 1 |
1 | 1107-1110 | Thus for points outside the shell, the field due
to a uniformly charged shell is as if the entire charge of the shell is
concentrated at its centre (ii) Field inside the shell: In Fig 1 28(b), the point P is inside the
shell |
1 | 1108-1111 | (ii) Field inside the shell: In Fig 1 28(b), the point P is inside the
shell The Gaussian surface is again a sphere through P centred at O |
1 | 1109-1112 | 1 28(b), the point P is inside the
shell The Gaussian surface is again a sphere through P centred at O FIGURE 1 |
1 | 1110-1113 | 28(b), the point P is inside the
shell The Gaussian surface is again a sphere through P centred at O FIGURE 1 28 Gaussian
surfaces for a point with
(a) r > R, (b) r < R |
1 | 1111-1114 | The Gaussian surface is again a sphere through P centred at O FIGURE 1 28 Gaussian
surfaces for a point with
(a) r > R, (b) r < R Rationalised 2023-24
36
Physics
EXAMPLE 1 |
1 | 1112-1115 | FIGURE 1 28 Gaussian
surfaces for a point with
(a) r > R, (b) r < R Rationalised 2023-24
36
Physics
EXAMPLE 1 12
The flux through the Gaussian surface, calculated as before, is
E × 4 p r2 |
1 | 1113-1116 | 28 Gaussian
surfaces for a point with
(a) r > R, (b) r < R Rationalised 2023-24
36
Physics
EXAMPLE 1 12
The flux through the Gaussian surface, calculated as before, is
E × 4 p r2 However, in this case, the Gaussian surface encloses no
charge |
1 | 1114-1117 | Rationalised 2023-24
36
Physics
EXAMPLE 1 12
The flux through the Gaussian surface, calculated as before, is
E × 4 p r2 However, in this case, the Gaussian surface encloses no
charge Gauss’s law then gives
E × 4 p r2 = 0
i |
1 | 1115-1118 | 12
The flux through the Gaussian surface, calculated as before, is
E × 4 p r2 However, in this case, the Gaussian surface encloses no
charge Gauss’s law then gives
E × 4 p r2 = 0
i e |
1 | 1116-1119 | However, in this case, the Gaussian surface encloses no
charge Gauss’s law then gives
E × 4 p r2 = 0
i e , E = 0 (r < R )
(1 |
1 | 1117-1120 | Gauss’s law then gives
E × 4 p r2 = 0
i e , E = 0 (r < R )
(1 35)
that is, the field due to a uniformly charged thin shell is zero at all points
inside the shell* |
1 | 1118-1121 | e , E = 0 (r < R )
(1 35)
that is, the field due to a uniformly charged thin shell is zero at all points
inside the shell* This important result is a direct consequence of Gauss’s
law which follows from Coulomb’s law |
1 | 1119-1122 | , E = 0 (r < R )
(1 35)
that is, the field due to a uniformly charged thin shell is zero at all points
inside the shell* This important result is a direct consequence of Gauss’s
law which follows from Coulomb’s law The experimental verification of
this result confirms the 1/r2 dependence in Coulomb’s law |
1 | 1120-1123 | 35)
that is, the field due to a uniformly charged thin shell is zero at all points
inside the shell* This important result is a direct consequence of Gauss’s
law which follows from Coulomb’s law The experimental verification of
this result confirms the 1/r2 dependence in Coulomb’s law Example 1 |
1 | 1121-1124 | This important result is a direct consequence of Gauss’s
law which follows from Coulomb’s law The experimental verification of
this result confirms the 1/r2 dependence in Coulomb’s law Example 1 12 An early model for an atom considered it to have a
positively charged point nucleus of charge Ze, surrounded by a
uniform density of negative charge up to a radius R |
1 | 1122-1125 | The experimental verification of
this result confirms the 1/r2 dependence in Coulomb’s law Example 1 12 An early model for an atom considered it to have a
positively charged point nucleus of charge Ze, surrounded by a
uniform density of negative charge up to a radius R The atom as a
whole is neutral |
1 | 1123-1126 | Example 1 12 An early model for an atom considered it to have a
positively charged point nucleus of charge Ze, surrounded by a
uniform density of negative charge up to a radius R The atom as a
whole is neutral For this model, what is the electric field at a distance
r from the nucleus |
1 | 1124-1127 | 12 An early model for an atom considered it to have a
positively charged point nucleus of charge Ze, surrounded by a
uniform density of negative charge up to a radius R The atom as a
whole is neutral For this model, what is the electric field at a distance
r from the nucleus FIGURE 1 |
1 | 1125-1128 | The atom as a
whole is neutral For this model, what is the electric field at a distance
r from the nucleus FIGURE 1 29
Solution The charge distribution for this model of the atom is as
shown in Fig |
1 | 1126-1129 | For this model, what is the electric field at a distance
r from the nucleus FIGURE 1 29
Solution The charge distribution for this model of the atom is as
shown in Fig 1 |
1 | 1127-1130 | FIGURE 1 29
Solution The charge distribution for this model of the atom is as
shown in Fig 1 29 |
1 | 1128-1131 | 29
Solution The charge distribution for this model of the atom is as
shown in Fig 1 29 The total negative charge in the uniform spherical
charge distribution of radius R must be –Z e, since the atom (nucleus
of charge Z e + negative charge) is neutral |
1 | 1129-1132 | 1 29 The total negative charge in the uniform spherical
charge distribution of radius R must be –Z e, since the atom (nucleus
of charge Z e + negative charge) is neutral This immediately gives us
the negative charge density r, since we must have
3
4
0–
3
R
Ze
ρ
π
=
or
3
3
4
Ze
R
ρ = −
π
To find the electric field E(r) at a point P which is a distance r away
from the nucleus, we use Gauss’s law |
1 | 1130-1133 | 29 The total negative charge in the uniform spherical
charge distribution of radius R must be –Z e, since the atom (nucleus
of charge Z e + negative charge) is neutral This immediately gives us
the negative charge density r, since we must have
3
4
0–
3
R
Ze
ρ
π
=
or
3
3
4
Ze
R
ρ = −
π
To find the electric field E(r) at a point P which is a distance r away
from the nucleus, we use Gauss’s law Because of the spherical
symmetry of the charge distribution, the magnitude of the electric
field E(r) depends only on the radial distance, no matter what the
direction of r |
1 | 1131-1134 | The total negative charge in the uniform spherical
charge distribution of radius R must be –Z e, since the atom (nucleus
of charge Z e + negative charge) is neutral This immediately gives us
the negative charge density r, since we must have
3
4
0–
3
R
Ze
ρ
π
=
or
3
3
4
Ze
R
ρ = −
π
To find the electric field E(r) at a point P which is a distance r away
from the nucleus, we use Gauss’s law Because of the spherical
symmetry of the charge distribution, the magnitude of the electric
field E(r) depends only on the radial distance, no matter what the
direction of r Its direction is along (or opposite to) the radius vector r
from the origin to the point P |
1 | 1132-1135 | This immediately gives us
the negative charge density r, since we must have
3
4
0–
3
R
Ze
ρ
π
=
or
3
3
4
Ze
R
ρ = −
π
To find the electric field E(r) at a point P which is a distance r away
from the nucleus, we use Gauss’s law Because of the spherical
symmetry of the charge distribution, the magnitude of the electric
field E(r) depends only on the radial distance, no matter what the
direction of r Its direction is along (or opposite to) the radius vector r
from the origin to the point P The obvious Gaussian surface is a
spherical surface centred at the nucleus |
1 | 1133-1136 | Because of the spherical
symmetry of the charge distribution, the magnitude of the electric
field E(r) depends only on the radial distance, no matter what the
direction of r Its direction is along (or opposite to) the radius vector r
from the origin to the point P The obvious Gaussian surface is a
spherical surface centred at the nucleus We consider two situations,
namely, r < R and r > R |
1 | 1134-1137 | Its direction is along (or opposite to) the radius vector r
from the origin to the point P The obvious Gaussian surface is a
spherical surface centred at the nucleus We consider two situations,
namely, r < R and r > R (i) r < R : The electric flux f enclosed by the spherical surface is
f = E (r) × 4 p r 2
*
Compare this with a uniform mass shell discussed in Section 7 |
1 | 1135-1138 | The obvious Gaussian surface is a
spherical surface centred at the nucleus We consider two situations,
namely, r < R and r > R (i) r < R : The electric flux f enclosed by the spherical surface is
f = E (r) × 4 p r 2
*
Compare this with a uniform mass shell discussed in Section 7 5 of Class XI
Textbook of Physics |
1 | 1136-1139 | We consider two situations,
namely, r < R and r > R (i) r < R : The electric flux f enclosed by the spherical surface is
f = E (r) × 4 p r 2
*
Compare this with a uniform mass shell discussed in Section 7 5 of Class XI
Textbook of Physics Rationalised 2023-24
Electric Charges
and Fields
37
EXAMPLE 1 |
1 | 1137-1140 | (i) r < R : The electric flux f enclosed by the spherical surface is
f = E (r) × 4 p r 2
*
Compare this with a uniform mass shell discussed in Section 7 5 of Class XI
Textbook of Physics Rationalised 2023-24
Electric Charges
and Fields
37
EXAMPLE 1 12
where E (r) is the magnitude of the electric field at r |
1 | 1138-1141 | 5 of Class XI
Textbook of Physics Rationalised 2023-24
Electric Charges
and Fields
37
EXAMPLE 1 12
where E (r) is the magnitude of the electric field at r This is because
the field at any point on the spherical Gaussian surface has the
same direction as the normal to the surface there, and has the same
magnitude at all points on the surface |
1 | 1139-1142 | Rationalised 2023-24
Electric Charges
and Fields
37
EXAMPLE 1 12
where E (r) is the magnitude of the electric field at r This is because
the field at any point on the spherical Gaussian surface has the
same direction as the normal to the surface there, and has the same
magnitude at all points on the surface The charge q enclosed by the Gaussian surface is the positive nuclear
charge and the negative charge within the sphere of radius r,
i |
1 | 1140-1143 | 12
where E (r) is the magnitude of the electric field at r This is because
the field at any point on the spherical Gaussian surface has the
same direction as the normal to the surface there, and has the same
magnitude at all points on the surface The charge q enclosed by the Gaussian surface is the positive nuclear
charge and the negative charge within the sphere of radius r,
i e |
1 | 1141-1144 | This is because
the field at any point on the spherical Gaussian surface has the
same direction as the normal to the surface there, and has the same
magnitude at all points on the surface The charge q enclosed by the Gaussian surface is the positive nuclear
charge and the negative charge within the sphere of radius r,
i e ,
3
4
3
r
q
Z e
ρ
π
=
+
Substituting for the charge density r obtained earlier, we have
3
3
r
q
Z e
Z e
R
=
−
Gauss’s law then gives,
2
3
0
1
( )
;
4
Z e
r
E r
r
R
r
R
ε
=
−
<
π
The electric field is directed radially outward |
1 | 1142-1145 | The charge q enclosed by the Gaussian surface is the positive nuclear
charge and the negative charge within the sphere of radius r,
i e ,
3
4
3
r
q
Z e
ρ
π
=
+
Substituting for the charge density r obtained earlier, we have
3
3
r
q
Z e
Z e
R
=
−
Gauss’s law then gives,
2
3
0
1
( )
;
4
Z e
r
E r
r
R
r
R
ε
=
−
<
π
The electric field is directed radially outward (ii) r > R: In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral |
1 | 1143-1146 | e ,
3
4
3
r
q
Z e
ρ
π
=
+
Substituting for the charge density r obtained earlier, we have
3
3
r
q
Z e
Z e
R
=
−
Gauss’s law then gives,
2
3
0
1
( )
;
4
Z e
r
E r
r
R
r
R
ε
=
−
<
π
The electric field is directed radially outward (ii) r > R: In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral Thus, from Gauss’s
law,
E (r) × 4 p r 2 = 0 or E (r) = 0; r > R
At r = R, both cases give the same result: E = 0 |
1 | 1144-1147 | ,
3
4
3
r
q
Z e
ρ
π
=
+
Substituting for the charge density r obtained earlier, we have
3
3
r
q
Z e
Z e
R
=
−
Gauss’s law then gives,
2
3
0
1
( )
;
4
Z e
r
E r
r
R
r
R
ε
=
−
<
π
The electric field is directed radially outward (ii) r > R: In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral Thus, from Gauss’s
law,
E (r) × 4 p r 2 = 0 or E (r) = 0; r > R
At r = R, both cases give the same result: E = 0 SUMMARY
1 |
1 | 1145-1148 | (ii) r > R: In this case, the total charge enclosed by the Gaussian
spherical surface is zero since the atom is neutral Thus, from Gauss’s
law,
E (r) × 4 p r 2 = 0 or E (r) = 0; r > R
At r = R, both cases give the same result: E = 0 SUMMARY
1 Electric and magnetic forces determine the properties of atoms,
molecules and bulk matter |
1 | 1146-1149 | Thus, from Gauss’s
law,
E (r) × 4 p r 2 = 0 or E (r) = 0; r > R
At r = R, both cases give the same result: E = 0 SUMMARY
1 Electric and magnetic forces determine the properties of atoms,
molecules and bulk matter 2 |
1 | 1147-1150 | SUMMARY
1 Electric and magnetic forces determine the properties of atoms,
molecules and bulk matter 2 From simple experiments on frictional electricity, one can infer that
there are two types of charges in nature; and that like charges repel
and unlike charges attract |
1 | 1148-1151 | Electric and magnetic forces determine the properties of atoms,
molecules and bulk matter 2 From simple experiments on frictional electricity, one can infer that
there are two types of charges in nature; and that like charges repel
and unlike charges attract By convention, the charge on a glass rod
rubbed with silk is positive; that on a plastic rod rubbed with fur is
then negative |
1 | 1149-1152 | 2 From simple experiments on frictional electricity, one can infer that
there are two types of charges in nature; and that like charges repel
and unlike charges attract By convention, the charge on a glass rod
rubbed with silk is positive; that on a plastic rod rubbed with fur is
then negative 3 |
1 | 1150-1153 | From simple experiments on frictional electricity, one can infer that
there are two types of charges in nature; and that like charges repel
and unlike charges attract By convention, the charge on a glass rod
rubbed with silk is positive; that on a plastic rod rubbed with fur is
then negative 3 Conductors allow movement of electric charge through them,
insulators do not |
1 | 1151-1154 | By convention, the charge on a glass rod
rubbed with silk is positive; that on a plastic rod rubbed with fur is
then negative 3 Conductors allow movement of electric charge through them,
insulators do not In metals, the mobile charges are electrons; in
electrolytes both positive and negative ions are mobile |
1 | 1152-1155 | 3 Conductors allow movement of electric charge through them,
insulators do not In metals, the mobile charges are electrons; in
electrolytes both positive and negative ions are mobile 4 |
1 | 1153-1156 | Conductors allow movement of electric charge through them,
insulators do not In metals, the mobile charges are electrons; in
electrolytes both positive and negative ions are mobile 4 Electric charge has three basic properties: quantisation, additivity
and conservation |
1 | 1154-1157 | In metals, the mobile charges are electrons; in
electrolytes both positive and negative ions are mobile 4 Electric charge has three basic properties: quantisation, additivity
and conservation Quantisation of electric charge means that total charge (q) of a body
is always an integral multiple of a basic quantum of charge (e) i |
1 | 1155-1158 | 4 Electric charge has three basic properties: quantisation, additivity
and conservation Quantisation of electric charge means that total charge (q) of a body
is always an integral multiple of a basic quantum of charge (e) i e |
1 | 1156-1159 | Electric charge has three basic properties: quantisation, additivity
and conservation Quantisation of electric charge means that total charge (q) of a body
is always an integral multiple of a basic quantum of charge (e) i e ,
q = n e, where n = 0, ±1, ±2, ±3, |
1 | 1157-1160 | Quantisation of electric charge means that total charge (q) of a body
is always an integral multiple of a basic quantum of charge (e) i e ,
q = n e, where n = 0, ±1, ±2, ±3, Proton and electron have charges
+e, –e, respectively |
1 | 1158-1161 | e ,
q = n e, where n = 0, ±1, ±2, ±3, Proton and electron have charges
+e, –e, respectively For macroscopic charges for which n is a very large
number, quantisation of charge can be ignored |
1 | 1159-1162 | ,
q = n e, where n = 0, ±1, ±2, ±3, Proton and electron have charges
+e, –e, respectively For macroscopic charges for which n is a very large
number, quantisation of charge can be ignored Additivity of electric charges means that the total charge of a system
is the algebraic sum (i |
1 | 1160-1163 | Proton and electron have charges
+e, –e, respectively For macroscopic charges for which n is a very large
number, quantisation of charge can be ignored Additivity of electric charges means that the total charge of a system
is the algebraic sum (i e |
1 | 1161-1164 | For macroscopic charges for which n is a very large
number, quantisation of charge can be ignored Additivity of electric charges means that the total charge of a system
is the algebraic sum (i e , the sum taking into account proper signs)
of all individual charges in the system |
1 | 1162-1165 | Additivity of electric charges means that the total charge of a system
is the algebraic sum (i e , the sum taking into account proper signs)
of all individual charges in the system Conservation of electric charges means that the total charge of an
isolated system remains unchanged with time |
1 | 1163-1166 | e , the sum taking into account proper signs)
of all individual charges in the system Conservation of electric charges means that the total charge of an
isolated system remains unchanged with time This means that when
Rationalised 2023-24
38
Physics
bodies are charged through friction, there is a transfer of electric charge
from one body to another, but no creation or destruction
of charge |
1 | 1164-1167 | , the sum taking into account proper signs)
of all individual charges in the system Conservation of electric charges means that the total charge of an
isolated system remains unchanged with time This means that when
Rationalised 2023-24
38
Physics
bodies are charged through friction, there is a transfer of electric charge
from one body to another, but no creation or destruction
of charge 5 |
1 | 1165-1168 | Conservation of electric charges means that the total charge of an
isolated system remains unchanged with time This means that when
Rationalised 2023-24
38
Physics
bodies are charged through friction, there is a transfer of electric charge
from one body to another, but no creation or destruction
of charge 5 Coulomb’s Law: The mutual electrostatic force between two point
charges q1 and q2 is proportional to the product q1q2 and inversely
proportional to the square of the distance r21 separating them |
1 | 1166-1169 | This means that when
Rationalised 2023-24
38
Physics
bodies are charged through friction, there is a transfer of electric charge
from one body to another, but no creation or destruction
of charge 5 Coulomb’s Law: The mutual electrostatic force between two point
charges q1 and q2 is proportional to the product q1q2 and inversely
proportional to the square of the distance r21 separating them Mathematically,
F21 = force on q2 due to
1
2
1
21
2
21
ˆ
k
(q q )
q
r
=
r
where
ˆr21
is a unit vector in the direction from q1 to q2 and k =
0
1
4 ε
π
is the constant of proportionality |
1 | 1167-1170 | 5 Coulomb’s Law: The mutual electrostatic force between two point
charges q1 and q2 is proportional to the product q1q2 and inversely
proportional to the square of the distance r21 separating them Mathematically,
F21 = force on q2 due to
1
2
1
21
2
21
ˆ
k
(q q )
q
r
=
r
where
ˆr21
is a unit vector in the direction from q1 to q2 and k =
0
1
4 ε
π
is the constant of proportionality In SI units, the unit of charge is coulomb |
1 | 1168-1171 | Coulomb’s Law: The mutual electrostatic force between two point
charges q1 and q2 is proportional to the product q1q2 and inversely
proportional to the square of the distance r21 separating them Mathematically,
F21 = force on q2 due to
1
2
1
21
2
21
ˆ
k
(q q )
q
r
=
r
where
ˆr21
is a unit vector in the direction from q1 to q2 and k =
0
1
4 ε
π
is the constant of proportionality In SI units, the unit of charge is coulomb The experimental value of
the constant e0 is
e0 = 8 |
1 | 1169-1172 | Mathematically,
F21 = force on q2 due to
1
2
1
21
2
21
ˆ
k
(q q )
q
r
=
r
where
ˆr21
is a unit vector in the direction from q1 to q2 and k =
0
1
4 ε
π
is the constant of proportionality In SI units, the unit of charge is coulomb The experimental value of
the constant e0 is
e0 = 8 854 × 10–12 C2 N–1 m–2
The approximate value of k is
k = 9 × 109 N m2 C–2
6 |
1 | 1170-1173 | In SI units, the unit of charge is coulomb The experimental value of
the constant e0 is
e0 = 8 854 × 10–12 C2 N–1 m–2
The approximate value of k is
k = 9 × 109 N m2 C–2
6 The ratio of electric force and gravitational force between a proton
and an electron is
2
39
2 4
10
e
p
k e |
1 | 1171-1174 | The experimental value of
the constant e0 is
e0 = 8 854 × 10–12 C2 N–1 m–2
The approximate value of k is
k = 9 × 109 N m2 C–2
6 The ratio of electric force and gravitational force between a proton
and an electron is
2
39
2 4
10
e
p
k e G
m m
≅
×
7 |
1 | 1172-1175 | 854 × 10–12 C2 N–1 m–2
The approximate value of k is
k = 9 × 109 N m2 C–2
6 The ratio of electric force and gravitational force between a proton
and an electron is
2
39
2 4
10
e
p
k e G
m m
≅
×
7 Superposition Principle: The principle is based on the property that the
forces with which two charges attract or repel each other are not
affected by the presence of a third (or more) additional charge(s) |
1 | 1173-1176 | The ratio of electric force and gravitational force between a proton
and an electron is
2
39
2 4
10
e
p
k e G
m m
≅
×
7 Superposition Principle: The principle is based on the property that the
forces with which two charges attract or repel each other are not
affected by the presence of a third (or more) additional charge(s) For
an assembly of charges q1, q2, q3, |
1 | 1174-1177 | G
m m
≅
×
7 Superposition Principle: The principle is based on the property that the
forces with which two charges attract or repel each other are not
affected by the presence of a third (or more) additional charge(s) For
an assembly of charges q1, q2, q3, , the force on any charge, say q1, is
the vector sum of the force on q1 due to q2, the force on q1 due to q3,
and so on |
1 | 1175-1178 | Superposition Principle: The principle is based on the property that the
forces with which two charges attract or repel each other are not
affected by the presence of a third (or more) additional charge(s) For
an assembly of charges q1, q2, q3, , the force on any charge, say q1, is
the vector sum of the force on q1 due to q2, the force on q1 due to q3,
and so on For each pair, the force is given by the Coulomb’s law for
two charges stated earlier |
1 | 1176-1179 | For
an assembly of charges q1, q2, q3, , the force on any charge, say q1, is
the vector sum of the force on q1 due to q2, the force on q1 due to q3,
and so on For each pair, the force is given by the Coulomb’s law for
two charges stated earlier 8 |
1 | 1177-1180 | , the force on any charge, say q1, is
the vector sum of the force on q1 due to q2, the force on q1 due to q3,
and so on For each pair, the force is given by the Coulomb’s law for
two charges stated earlier 8 The electric field E at a point due to a charge configuration is the
force on a small positive test charge q placed at the point divided by
the magnitude of the charge |
1 | 1178-1181 | For each pair, the force is given by the Coulomb’s law for
two charges stated earlier 8 The electric field E at a point due to a charge configuration is the
force on a small positive test charge q placed at the point divided by
the magnitude of the charge Electric field due to a point charge q has
a magnitude |q|/4pe0r2; it is radially outwards from q, if q is positive,
and radially inwards if q is negative |
1 | 1179-1182 | 8 The electric field E at a point due to a charge configuration is the
force on a small positive test charge q placed at the point divided by
the magnitude of the charge Electric field due to a point charge q has
a magnitude |q|/4pe0r2; it is radially outwards from q, if q is positive,
and radially inwards if q is negative Like Coulomb force, electric field
also satisfies superposition principle |
1 | 1180-1183 | The electric field E at a point due to a charge configuration is the
force on a small positive test charge q placed at the point divided by
the magnitude of the charge Electric field due to a point charge q has
a magnitude |q|/4pe0r2; it is radially outwards from q, if q is positive,
and radially inwards if q is negative Like Coulomb force, electric field
also satisfies superposition principle 9 |
1 | 1181-1184 | Electric field due to a point charge q has
a magnitude |q|/4pe0r2; it is radially outwards from q, if q is positive,
and radially inwards if q is negative Like Coulomb force, electric field
also satisfies superposition principle 9 An electric field line is a curve drawn in such a way that the tangent
at each point on the curve gives the direction of electric field at that
point |
1 | 1182-1185 | Like Coulomb force, electric field
also satisfies superposition principle 9 An electric field line is a curve drawn in such a way that the tangent
at each point on the curve gives the direction of electric field at that
point The relative closeness of field lines indicates the relative strength
of electric field at different points; they crowd near each other in regions
of strong electric field and are far apart where the electric field is
weak |
1 | 1183-1186 | 9 An electric field line is a curve drawn in such a way that the tangent
at each point on the curve gives the direction of electric field at that
point The relative closeness of field lines indicates the relative strength
of electric field at different points; they crowd near each other in regions
of strong electric field and are far apart where the electric field is
weak In regions of constant electric field, the field lines are uniformly
spaced parallel straight lines |
1 | 1184-1187 | An electric field line is a curve drawn in such a way that the tangent
at each point on the curve gives the direction of electric field at that
point The relative closeness of field lines indicates the relative strength
of electric field at different points; they crowd near each other in regions
of strong electric field and are far apart where the electric field is
weak In regions of constant electric field, the field lines are uniformly
spaced parallel straight lines 10 |
1 | 1185-1188 | The relative closeness of field lines indicates the relative strength
of electric field at different points; they crowd near each other in regions
of strong electric field and are far apart where the electric field is
weak In regions of constant electric field, the field lines are uniformly
spaced parallel straight lines 10 Some of the important properties of field lines are: (i) Field lines are
continuous curves without any breaks |
1 | 1186-1189 | In regions of constant electric field, the field lines are uniformly
spaced parallel straight lines 10 Some of the important properties of field lines are: (i) Field lines are
continuous curves without any breaks (ii) Two field lines cannot cross
each other |
1 | 1187-1190 | 10 Some of the important properties of field lines are: (i) Field lines are
continuous curves without any breaks (ii) Two field lines cannot cross
each other (iii) Electrostatic field lines start at positive charges and
end at negative charges —they cannot form closed loops |
1 | 1188-1191 | Some of the important properties of field lines are: (i) Field lines are
continuous curves without any breaks (ii) Two field lines cannot cross
each other (iii) Electrostatic field lines start at positive charges and
end at negative charges —they cannot form closed loops 11 |
1 | 1189-1192 | (ii) Two field lines cannot cross
each other (iii) Electrostatic field lines start at positive charges and
end at negative charges —they cannot form closed loops 11 An electric dipole is a pair of equal and opposite charges q and –q
separated by some distance 2a |
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