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1 | 290-293 | The value for acceleration of the proton is
2 3 × 10–8 N / 1 67 × 10–27 kg = 1 4 × 1019 m/s2
EXAMPLE 1 |
1 | 291-294 | 3 × 10–8 N / 1 67 × 10–27 kg = 1 4 × 1019 m/s2
EXAMPLE 1 3
FIGURE 1 |
1 | 292-295 | 67 × 10–27 kg = 1 4 × 1019 m/s2
EXAMPLE 1 3
FIGURE 1 4
Example 1 |
1 | 293-296 | 4 × 1019 m/s2
EXAMPLE 1 3
FIGURE 1 4
Example 1 4 A charged metallic sphere A is suspended by a nylon
thread |
1 | 294-297 | 3
FIGURE 1 4
Example 1 4 A charged metallic sphere A is suspended by a nylon
thread Another charged metallic sphere B held by an insulating
EXAMPLE 1 |
1 | 295-298 | 4
Example 1 4 A charged metallic sphere A is suspended by a nylon
thread Another charged metallic sphere B held by an insulating
EXAMPLE 1 4
Rationalised 2023-24
Electric Charges
and Fields
11
EXAMPLE 1 |
1 | 296-299 | 4 A charged metallic sphere A is suspended by a nylon
thread Another charged metallic sphere B held by an insulating
EXAMPLE 1 4
Rationalised 2023-24
Electric Charges
and Fields
11
EXAMPLE 1 4
handle is brought close to A such that the distance between their
centres is 10 cm, as shown in Fig |
1 | 297-300 | Another charged metallic sphere B held by an insulating
EXAMPLE 1 4
Rationalised 2023-24
Electric Charges
and Fields
11
EXAMPLE 1 4
handle is brought close to A such that the distance between their
centres is 10 cm, as shown in Fig 1 |
1 | 298-301 | 4
Rationalised 2023-24
Electric Charges
and Fields
11
EXAMPLE 1 4
handle is brought close to A such that the distance between their
centres is 10 cm, as shown in Fig 1 4(a) |
1 | 299-302 | 4
handle is brought close to A such that the distance between their
centres is 10 cm, as shown in Fig 1 4(a) The resulting repulsion of A
is noted (for example, by shining a beam of light and measuring the
deflection of its shadow on a screen) |
1 | 300-303 | 1 4(a) The resulting repulsion of A
is noted (for example, by shining a beam of light and measuring the
deflection of its shadow on a screen) Spheres A and B are touched
by uncharged spheres C and D respectively, as shown in Fig |
1 | 301-304 | 4(a) The resulting repulsion of A
is noted (for example, by shining a beam of light and measuring the
deflection of its shadow on a screen) Spheres A and B are touched
by uncharged spheres C and D respectively, as shown in Fig 1 |
1 | 302-305 | The resulting repulsion of A
is noted (for example, by shining a beam of light and measuring the
deflection of its shadow on a screen) Spheres A and B are touched
by uncharged spheres C and D respectively, as shown in Fig 1 4(b) |
1 | 303-306 | Spheres A and B are touched
by uncharged spheres C and D respectively, as shown in Fig 1 4(b) C and D are then removed and B is brought closer to A to a
distance of 5 |
1 | 304-307 | 1 4(b) C and D are then removed and B is brought closer to A to a
distance of 5 0 cm between their centres, as shown in Fig |
1 | 305-308 | 4(b) C and D are then removed and B is brought closer to A to a
distance of 5 0 cm between their centres, as shown in Fig 1 |
1 | 306-309 | C and D are then removed and B is brought closer to A to a
distance of 5 0 cm between their centres, as shown in Fig 1 4(c) |
1 | 307-310 | 0 cm between their centres, as shown in Fig 1 4(c) What is the expected repulsion of A on the basis of Coulomb’s law |
1 | 308-311 | 1 4(c) What is the expected repulsion of A on the basis of Coulomb’s law Spheres A and C and spheres B and D have identical sizes |
1 | 309-312 | 4(c) What is the expected repulsion of A on the basis of Coulomb’s law Spheres A and C and spheres B and D have identical sizes Ignore
the sizes of A and B in comparison to the separation between their
centres |
1 | 310-313 | What is the expected repulsion of A on the basis of Coulomb’s law Spheres A and C and spheres B and D have identical sizes Ignore
the sizes of A and B in comparison to the separation between their
centres Solution Let the original charge on sphere A be q and that on B be
q¢ |
1 | 311-314 | Spheres A and C and spheres B and D have identical sizes Ignore
the sizes of A and B in comparison to the separation between their
centres Solution Let the original charge on sphere A be q and that on B be
q¢ At a distance r between their centres, the magnitude of the
electrostatic force on each is given by
F
rqq
=
′
41
0
2
πε
neglecting the sizes of spheres A and B in comparison to r |
1 | 312-315 | Ignore
the sizes of A and B in comparison to the separation between their
centres Solution Let the original charge on sphere A be q and that on B be
q¢ At a distance r between their centres, the magnitude of the
electrostatic force on each is given by
F
rqq
=
′
41
0
2
πε
neglecting the sizes of spheres A and B in comparison to r When an
identical but uncharged sphere C touches A, the charges redistribute
on A and C and, by symmetry, each sphere carries a charge q/2 |
1 | 313-316 | Solution Let the original charge on sphere A be q and that on B be
q¢ At a distance r between their centres, the magnitude of the
electrostatic force on each is given by
F
rqq
=
′
41
0
2
πε
neglecting the sizes of spheres A and B in comparison to r When an
identical but uncharged sphere C touches A, the charges redistribute
on A and C and, by symmetry, each sphere carries a charge q/2 Similarly, after D touches B, the redistributed charge on each is
q¢/2 |
1 | 314-317 | At a distance r between their centres, the magnitude of the
electrostatic force on each is given by
F
rqq
=
′
41
0
2
πε
neglecting the sizes of spheres A and B in comparison to r When an
identical but uncharged sphere C touches A, the charges redistribute
on A and C and, by symmetry, each sphere carries a charge q/2 Similarly, after D touches B, the redistributed charge on each is
q¢/2 Now, if the separation between A and B is halved, the magnitude
of the electrostatic force on each is
′ =
′
=
′ =
F
q
q
r
rqq
F
41
2
2
2
41
0
2
0
2
π
π
ε
ε
( / )(
( / )/ )
(
)
Thus the electrostatic force on A, due to B, remains unaltered |
1 | 315-318 | When an
identical but uncharged sphere C touches A, the charges redistribute
on A and C and, by symmetry, each sphere carries a charge q/2 Similarly, after D touches B, the redistributed charge on each is
q¢/2 Now, if the separation between A and B is halved, the magnitude
of the electrostatic force on each is
′ =
′
=
′ =
F
q
q
r
rqq
F
41
2
2
2
41
0
2
0
2
π
π
ε
ε
( / )(
( / )/ )
(
)
Thus the electrostatic force on A, due to B, remains unaltered 1 |
1 | 316-319 | Similarly, after D touches B, the redistributed charge on each is
q¢/2 Now, if the separation between A and B is halved, the magnitude
of the electrostatic force on each is
′ =
′
=
′ =
F
q
q
r
rqq
F
41
2
2
2
41
0
2
0
2
π
π
ε
ε
( / )(
( / )/ )
(
)
Thus the electrostatic force on A, due to B, remains unaltered 1 6 FORCES BETWEEN MULTIPLE CHARGES
The mutual electric force between two charges is given by
Coulomb’s law |
1 | 317-320 | Now, if the separation between A and B is halved, the magnitude
of the electrostatic force on each is
′ =
′
=
′ =
F
q
q
r
rqq
F
41
2
2
2
41
0
2
0
2
π
π
ε
ε
( / )(
( / )/ )
(
)
Thus the electrostatic force on A, due to B, remains unaltered 1 6 FORCES BETWEEN MULTIPLE CHARGES
The mutual electric force between two charges is given by
Coulomb’s law How to calculate the force on a charge where
there are not one but several charges around |
1 | 318-321 | 1 6 FORCES BETWEEN MULTIPLE CHARGES
The mutual electric force between two charges is given by
Coulomb’s law How to calculate the force on a charge where
there are not one but several charges around Consider a
system of n stationary charges q1, q2, q3, |
1 | 319-322 | 6 FORCES BETWEEN MULTIPLE CHARGES
The mutual electric force between two charges is given by
Coulomb’s law How to calculate the force on a charge where
there are not one but several charges around Consider a
system of n stationary charges q1, q2, q3, , qn in vacuum |
1 | 320-323 | How to calculate the force on a charge where
there are not one but several charges around Consider a
system of n stationary charges q1, q2, q3, , qn in vacuum What is the force on q1 due to q2, q3, |
1 | 321-324 | Consider a
system of n stationary charges q1, q2, q3, , qn in vacuum What is the force on q1 due to q2, q3, , qn |
1 | 322-325 | , qn in vacuum What is the force on q1 due to q2, q3, , qn Coulomb’s law is
not enough to answer this question |
1 | 323-326 | What is the force on q1 due to q2, q3, , qn Coulomb’s law is
not enough to answer this question Recall that forces of
mechanical origin add according to the parallelogram law of
addition |
1 | 324-327 | , qn Coulomb’s law is
not enough to answer this question Recall that forces of
mechanical origin add according to the parallelogram law of
addition Is the same true for forces of electrostatic origin |
1 | 325-328 | Coulomb’s law is
not enough to answer this question Recall that forces of
mechanical origin add according to the parallelogram law of
addition Is the same true for forces of electrostatic origin Experimentally, it is verified that force on any charge due
to a number of other charges is the vector sum of all the forces
on that charge due to the other charges, taken one at a time |
1 | 326-329 | Recall that forces of
mechanical origin add according to the parallelogram law of
addition Is the same true for forces of electrostatic origin Experimentally, it is verified that force on any charge due
to a number of other charges is the vector sum of all the forces
on that charge due to the other charges, taken one at a time The individual forces are unaffected due to the presence of
other charges |
1 | 327-330 | Is the same true for forces of electrostatic origin Experimentally, it is verified that force on any charge due
to a number of other charges is the vector sum of all the forces
on that charge due to the other charges, taken one at a time The individual forces are unaffected due to the presence of
other charges This is termed as the principle of superposition |
1 | 328-331 | Experimentally, it is verified that force on any charge due
to a number of other charges is the vector sum of all the forces
on that charge due to the other charges, taken one at a time The individual forces are unaffected due to the presence of
other charges This is termed as the principle of superposition To better understand the concept, consider a system of
three charges q1, q2 and q3, as shown in Fig |
1 | 329-332 | The individual forces are unaffected due to the presence of
other charges This is termed as the principle of superposition To better understand the concept, consider a system of
three charges q1, q2 and q3, as shown in Fig 1 |
1 | 330-333 | This is termed as the principle of superposition To better understand the concept, consider a system of
three charges q1, q2 and q3, as shown in Fig 1 5(a) |
1 | 331-334 | To better understand the concept, consider a system of
three charges q1, q2 and q3, as shown in Fig 1 5(a) The force
on one charge, say q1, due to two other charges q2, q3 can
therefore be obtained by performing a vector addition of the
forces due to each one of these charges |
1 | 332-335 | 1 5(a) The force
on one charge, say q1, due to two other charges q2, q3 can
therefore be obtained by performing a vector addition of the
forces due to each one of these charges Thus, if the force on q1
due to q2 is denoted by F12, F12 is given by Eq |
1 | 333-336 | 5(a) The force
on one charge, say q1, due to two other charges q2, q3 can
therefore be obtained by performing a vector addition of the
forces due to each one of these charges Thus, if the force on q1
due to q2 is denoted by F12, F12 is given by Eq (1 |
1 | 334-337 | The force
on one charge, say q1, due to two other charges q2, q3 can
therefore be obtained by performing a vector addition of the
forces due to each one of these charges Thus, if the force on q1
due to q2 is denoted by F12, F12 is given by Eq (1 3) even
though other charges are present |
1 | 335-338 | Thus, if the force on q1
due to q2 is denoted by F12, F12 is given by Eq (1 3) even
though other charges are present Thus,
F12 =
41
0
1
2
12
2
12
πε
rq q
ˆr
FIGURE 1 |
1 | 336-339 | (1 3) even
though other charges are present Thus,
F12 =
41
0
1
2
12
2
12
πε
rq q
ˆr
FIGURE 1 5 A system of
(a) three charges
(b) multiple charges |
1 | 337-340 | 3) even
though other charges are present Thus,
F12 =
41
0
1
2
12
2
12
πε
rq q
ˆr
FIGURE 1 5 A system of
(a) three charges
(b) multiple charges Rationalised 2023-24
12
Physics
EXAMPLE 1 |
1 | 338-341 | Thus,
F12 =
41
0
1
2
12
2
12
πε
rq q
ˆr
FIGURE 1 5 A system of
(a) three charges
(b) multiple charges Rationalised 2023-24
12
Physics
EXAMPLE 1 5
In the same way, the force on q1 due to q3, denoted by F13, is given by
F
r
13
0
1
3
13
2
13
=41
πε
rq q
ˆ
which again is the Coulomb force on q1 due to q3, even though other
charge q2 is present |
1 | 339-342 | 5 A system of
(a) three charges
(b) multiple charges Rationalised 2023-24
12
Physics
EXAMPLE 1 5
In the same way, the force on q1 due to q3, denoted by F13, is given by
F
r
13
0
1
3
13
2
13
=41
πε
rq q
ˆ
which again is the Coulomb force on q1 due to q3, even though other
charge q2 is present Thus the total force F1 on q1 due to the two charges q2 and q3 is
given as
F
F
F
r
r
1
12
13
0
1
2
12
2
12
0
1
3
13
2
13
41
41
=
+
=
+
π
π
ε
ε
rq q
rq q
ˆ
ˆ
(1 |
1 | 340-343 | Rationalised 2023-24
12
Physics
EXAMPLE 1 5
In the same way, the force on q1 due to q3, denoted by F13, is given by
F
r
13
0
1
3
13
2
13
=41
πε
rq q
ˆ
which again is the Coulomb force on q1 due to q3, even though other
charge q2 is present Thus the total force F1 on q1 due to the two charges q2 and q3 is
given as
F
F
F
r
r
1
12
13
0
1
2
12
2
12
0
1
3
13
2
13
41
41
=
+
=
+
π
π
ε
ε
rq q
rq q
ˆ
ˆ
(1 4)
The above calculation of force can be generalised to a system of
charges more than three, as shown in Fig |
1 | 341-344 | 5
In the same way, the force on q1 due to q3, denoted by F13, is given by
F
r
13
0
1
3
13
2
13
=41
πε
rq q
ˆ
which again is the Coulomb force on q1 due to q3, even though other
charge q2 is present Thus the total force F1 on q1 due to the two charges q2 and q3 is
given as
F
F
F
r
r
1
12
13
0
1
2
12
2
12
0
1
3
13
2
13
41
41
=
+
=
+
π
π
ε
ε
rq q
rq q
ˆ
ˆ
(1 4)
The above calculation of force can be generalised to a system of
charges more than three, as shown in Fig 1 |
1 | 342-345 | Thus the total force F1 on q1 due to the two charges q2 and q3 is
given as
F
F
F
r
r
1
12
13
0
1
2
12
2
12
0
1
3
13
2
13
41
41
=
+
=
+
π
π
ε
ε
rq q
rq q
ˆ
ˆ
(1 4)
The above calculation of force can be generalised to a system of
charges more than three, as shown in Fig 1 5(b) |
1 | 343-346 | 4)
The above calculation of force can be generalised to a system of
charges more than three, as shown in Fig 1 5(b) The principle of superposition says that in a system of charges q1,
q2, |
1 | 344-347 | 1 5(b) The principle of superposition says that in a system of charges q1,
q2, , qn, the force on q1 due to q2 is the same as given by Coulomb’s law,
i |
1 | 345-348 | 5(b) The principle of superposition says that in a system of charges q1,
q2, , qn, the force on q1 due to q2 is the same as given by Coulomb’s law,
i e |
1 | 346-349 | The principle of superposition says that in a system of charges q1,
q2, , qn, the force on q1 due to q2 is the same as given by Coulomb’s law,
i e , it is unaffected by the presence of the other charges q3, q4, |
1 | 347-350 | , qn, the force on q1 due to q2 is the same as given by Coulomb’s law,
i e , it is unaffected by the presence of the other charges q3, q4, , qn |
1 | 348-351 | e , it is unaffected by the presence of the other charges q3, q4, , qn The
total force F1 on the charge q1, due to all other charges, is then given by
the vector sum of the forces F12, F13, |
1 | 349-352 | , it is unaffected by the presence of the other charges q3, q4, , qn The
total force F1 on the charge q1, due to all other charges, is then given by
the vector sum of the forces F12, F13, , F1n:
i |
1 | 350-353 | , qn The
total force F1 on the charge q1, due to all other charges, is then given by
the vector sum of the forces F12, F13, , F1n:
i e |
1 | 351-354 | The
total force F1 on the charge q1, due to all other charges, is then given by
the vector sum of the forces F12, F13, , F1n:
i e ,
F
F
F
F
r
r
1
12
13
1n
=
+
+ |
1 | 352-355 | , F1n:
i e ,
F
F
F
F
r
r
1
12
13
1n
=
+
+ +
=
+
41
0
1
2
12
2
12
1
3
13
2
13
πε
rq q
rq q
ˆ
ˆ
+
+
|
1 | 353-356 | e ,
F
F
F
F
r
r
1
12
13
1n
=
+
+ +
=
+
41
0
1
2
12
2
12
1
3
13
2
13
πε
rq q
rq q
ˆ
ˆ
+
+
ˆ
q q
r
n
n
n
1
1
2
1r
=
=∑
q
q
r
i
i
i
n
i
1
0
1
2
2
1
4πε
ˆr
(1 |
1 | 354-357 | ,
F
F
F
F
r
r
1
12
13
1n
=
+
+ +
=
+
41
0
1
2
12
2
12
1
3
13
2
13
πε
rq q
rq q
ˆ
ˆ
+
+
ˆ
q q
r
n
n
n
1
1
2
1r
=
=∑
q
q
r
i
i
i
n
i
1
0
1
2
2
1
4πε
ˆr
(1 5)
The vector sum is obtained as usual by the parallelogram law of
addition of vectors |
1 | 355-358 | +
=
+
41
0
1
2
12
2
12
1
3
13
2
13
πε
rq q
rq q
ˆ
ˆ
+
+
ˆ
q q
r
n
n
n
1
1
2
1r
=
=∑
q
q
r
i
i
i
n
i
1
0
1
2
2
1
4πε
ˆr
(1 5)
The vector sum is obtained as usual by the parallelogram law of
addition of vectors All of electrostatics is basically a consequence of
Coulomb’s law and the superposition principle |
1 | 356-359 | ˆ
q q
r
n
n
n
1
1
2
1r
=
=∑
q
q
r
i
i
i
n
i
1
0
1
2
2
1
4πε
ˆr
(1 5)
The vector sum is obtained as usual by the parallelogram law of
addition of vectors All of electrostatics is basically a consequence of
Coulomb’s law and the superposition principle Example 1 |
1 | 357-360 | 5)
The vector sum is obtained as usual by the parallelogram law of
addition of vectors All of electrostatics is basically a consequence of
Coulomb’s law and the superposition principle Example 1 5 Consider three charges q1, q2, q3 each equal to q at the
vertices of an equilateral triangle of side l |
1 | 358-361 | All of electrostatics is basically a consequence of
Coulomb’s law and the superposition principle Example 1 5 Consider three charges q1, q2, q3 each equal to q at the
vertices of an equilateral triangle of side l What is the force on a
charge Q (with the same sign as q) placed at the centroid of the
triangle, as shown in Fig |
1 | 359-362 | Example 1 5 Consider three charges q1, q2, q3 each equal to q at the
vertices of an equilateral triangle of side l What is the force on a
charge Q (with the same sign as q) placed at the centroid of the
triangle, as shown in Fig 1 |
1 | 360-363 | 5 Consider three charges q1, q2, q3 each equal to q at the
vertices of an equilateral triangle of side l What is the force on a
charge Q (with the same sign as q) placed at the centroid of the
triangle, as shown in Fig 1 6 |
1 | 361-364 | What is the force on a
charge Q (with the same sign as q) placed at the centroid of the
triangle, as shown in Fig 1 6 FIGURE 1 |
1 | 362-365 | 1 6 FIGURE 1 6
Solution In the given equilateral triangle ABC of sides of length l, if
we draw a perpendicular AD to the side BC,
AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O
from A is (2/3) AD = (1/ 3 ) l |
1 | 363-366 | 6 FIGURE 1 6
Solution In the given equilateral triangle ABC of sides of length l, if
we draw a perpendicular AD to the side BC,
AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O
from A is (2/3) AD = (1/ 3 ) l By symmatry AO = BO = CO |
1 | 364-367 | FIGURE 1 6
Solution In the given equilateral triangle ABC of sides of length l, if
we draw a perpendicular AD to the side BC,
AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O
from A is (2/3) AD = (1/ 3 ) l By symmatry AO = BO = CO Rationalised 2023-24
Electric Charges
and Fields
13
EXAMPLE 1 |
1 | 365-368 | 6
Solution In the given equilateral triangle ABC of sides of length l, if
we draw a perpendicular AD to the side BC,
AD = AC cos 30º = ( 3 /2 ) l and the distance AO of the centroid O
from A is (2/3) AD = (1/ 3 ) l By symmatry AO = BO = CO Rationalised 2023-24
Electric Charges
and Fields
13
EXAMPLE 1 5
Thus,
Force F1 on Q due to charge q at A =
43
0
2
πε
lQq
along AO
Force F2 on Q due to charge q at B =
43
0
2
πε
lQq
along BO
Force F3 on Q due to charge q at C =
43
0
2
πε
lQq
along CO
The resultant of forces F2 and F3 is
43
0
2
πε
lQq
along OA, by the
parallelogram law |
1 | 366-369 | By symmatry AO = BO = CO Rationalised 2023-24
Electric Charges
and Fields
13
EXAMPLE 1 5
Thus,
Force F1 on Q due to charge q at A =
43
0
2
πε
lQq
along AO
Force F2 on Q due to charge q at B =
43
0
2
πε
lQq
along BO
Force F3 on Q due to charge q at C =
43
0
2
πε
lQq
along CO
The resultant of forces F2 and F3 is
43
0
2
πε
lQq
along OA, by the
parallelogram law Therefore, the total force on Q =
43
0
2
πε
lQq
ˆ
ˆ
r
(−r
)
= 0, where ˆr
is the unit vector along OA |
1 | 367-370 | Rationalised 2023-24
Electric Charges
and Fields
13
EXAMPLE 1 5
Thus,
Force F1 on Q due to charge q at A =
43
0
2
πε
lQq
along AO
Force F2 on Q due to charge q at B =
43
0
2
πε
lQq
along BO
Force F3 on Q due to charge q at C =
43
0
2
πε
lQq
along CO
The resultant of forces F2 and F3 is
43
0
2
πε
lQq
along OA, by the
parallelogram law Therefore, the total force on Q =
43
0
2
πε
lQq
ˆ
ˆ
r
(−r
)
= 0, where ˆr
is the unit vector along OA It is clear also by symmetry that the three forces will sum to zero |
1 | 368-371 | 5
Thus,
Force F1 on Q due to charge q at A =
43
0
2
πε
lQq
along AO
Force F2 on Q due to charge q at B =
43
0
2
πε
lQq
along BO
Force F3 on Q due to charge q at C =
43
0
2
πε
lQq
along CO
The resultant of forces F2 and F3 is
43
0
2
πε
lQq
along OA, by the
parallelogram law Therefore, the total force on Q =
43
0
2
πε
lQq
ˆ
ˆ
r
(−r
)
= 0, where ˆr
is the unit vector along OA It is clear also by symmetry that the three forces will sum to zero Suppose that the resultant force was non-zero but in some direction |
1 | 369-372 | Therefore, the total force on Q =
43
0
2
πε
lQq
ˆ
ˆ
r
(−r
)
= 0, where ˆr
is the unit vector along OA It is clear also by symmetry that the three forces will sum to zero Suppose that the resultant force was non-zero but in some direction Consider what would happen if the system was rotated through 60°
about O |
1 | 370-373 | It is clear also by symmetry that the three forces will sum to zero Suppose that the resultant force was non-zero but in some direction Consider what would happen if the system was rotated through 60°
about O Example 1 |
1 | 371-374 | Suppose that the resultant force was non-zero but in some direction Consider what would happen if the system was rotated through 60°
about O Example 1 6 Consider the charges q, q, and –q placed at the vertices
of an equilateral triangle, as shown in Fig |
1 | 372-375 | Consider what would happen if the system was rotated through 60°
about O Example 1 6 Consider the charges q, q, and –q placed at the vertices
of an equilateral triangle, as shown in Fig 1 |
1 | 373-376 | Example 1 6 Consider the charges q, q, and –q placed at the vertices
of an equilateral triangle, as shown in Fig 1 7 |
1 | 374-377 | 6 Consider the charges q, q, and –q placed at the vertices
of an equilateral triangle, as shown in Fig 1 7 What is the force on
each charge |
1 | 375-378 | 1 7 What is the force on
each charge FIGURE 1 |
1 | 376-379 | 7 What is the force on
each charge FIGURE 1 7
Solution The forces acting on charge q at A due to charges q at B
and –q at C are F12 along BA and F13 along AC respectively, as shown
in Fig |
1 | 377-380 | What is the force on
each charge FIGURE 1 7
Solution The forces acting on charge q at A due to charges q at B
and –q at C are F12 along BA and F13 along AC respectively, as shown
in Fig 1 |
1 | 378-381 | FIGURE 1 7
Solution The forces acting on charge q at A due to charges q at B
and –q at C are F12 along BA and F13 along AC respectively, as shown
in Fig 1 7 |
1 | 379-382 | 7
Solution The forces acting on charge q at A due to charges q at B
and –q at C are F12 along BA and F13 along AC respectively, as shown
in Fig 1 7 By the parallelogram law, the total force F1 on the charge
q at A is given by
F1 = F
1ˆr where
1ˆr is a unit vector along BC |
1 | 380-383 | 1 7 By the parallelogram law, the total force F1 on the charge
q at A is given by
F1 = F
1ˆr where
1ˆr is a unit vector along BC The force of attraction or repulsion for each pair of charges has the
same magnitude F
q
=
2
0
2
4 π ε l
The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a
unit vector along AC |
1 | 381-384 | 7 By the parallelogram law, the total force F1 on the charge
q at A is given by
F1 = F
1ˆr where
1ˆr is a unit vector along BC The force of attraction or repulsion for each pair of charges has the
same magnitude F
q
=
2
0
2
4 π ε l
The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a
unit vector along AC EXAMPLE 1 |
1 | 382-385 | By the parallelogram law, the total force F1 on the charge
q at A is given by
F1 = F
1ˆr where
1ˆr is a unit vector along BC The force of attraction or repulsion for each pair of charges has the
same magnitude F
q
=
2
0
2
4 π ε l
The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a
unit vector along AC EXAMPLE 1 6
Rationalised 2023-24
14
Physics
EXAMPLE 1 |
1 | 383-386 | The force of attraction or repulsion for each pair of charges has the
same magnitude F
q
=
2
0
2
4 π ε l
The total force F2 on charge q at B is thus F2 = F ˆr 2, where ˆr 2 is a
unit vector along AC EXAMPLE 1 6
Rationalised 2023-24
14
Physics
EXAMPLE 1 6
Similarly the total force on charge –q at C is F3 =
3 F ˆn , where ˆn is
the unit vector along the direction bisecting the ÐBCA |
1 | 384-387 | EXAMPLE 1 6
Rationalised 2023-24
14
Physics
EXAMPLE 1 6
Similarly the total force on charge –q at C is F3 =
3 F ˆn , where ˆn is
the unit vector along the direction bisecting the ÐBCA It is interesting to see that the sum of the forces on the three charges
is zero, i |
1 | 385-388 | 6
Rationalised 2023-24
14
Physics
EXAMPLE 1 6
Similarly the total force on charge –q at C is F3 =
3 F ˆn , where ˆn is
the unit vector along the direction bisecting the ÐBCA It is interesting to see that the sum of the forces on the three charges
is zero, i e |
1 | 386-389 | 6
Similarly the total force on charge –q at C is F3 =
3 F ˆn , where ˆn is
the unit vector along the direction bisecting the ÐBCA It is interesting to see that the sum of the forces on the three charges
is zero, i e ,
F1 + F2 + F3 = 0
The result is not at all surprising |
1 | 387-390 | It is interesting to see that the sum of the forces on the three charges
is zero, i e ,
F1 + F2 + F3 = 0
The result is not at all surprising It follows straight from the fact
that Coulomb’s law is consistent with Newton’s third law |
1 | 388-391 | e ,
F1 + F2 + F3 = 0
The result is not at all surprising It follows straight from the fact
that Coulomb’s law is consistent with Newton’s third law The proof
is left to you as an exercise |
1 | 389-392 | ,
F1 + F2 + F3 = 0
The result is not at all surprising It follows straight from the fact
that Coulomb’s law is consistent with Newton’s third law The proof
is left to you as an exercise 1 |
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