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Chapter Sixteen: Problems
Note: These problems come from "Fundamentals of Senior Physics" by Parham and Webber.
1.
Two like charges experience a force F upon each other when placed a fixed distance apart in
air. What is the magnitude of the force if the distance between the two charges is:
(a)
doubled
d×2 ⇒ F/4
(b)
halved?
d×1/2 ⇒ F×4
What is the magnitude of the force if the magnitude of one of the charges is:
(a)
doubled
q×2 ⇒ F×2
(b)
halved
q×1/2 ⇒ F×1/2
2.
How much must you alter the distance between two charged objects in order to keep the force
between them constant, if you also:
(a)
treble the net charge on each
q1×3 & q2×3 ⇒ F×9
(b)
halve the net charge on each
q1×1/2 & q2×1/2 ⇒ F×1/4
(c)
want F×4 ⇒ d×1/2
double the charge on one and halve the charge on the other?
q1×2 & q2×1/2 ⇒ F×1
3.
want F×1/9 ⇒ d×3
want F×1 ⇒ d×1 (no change)
A positive test charge is placed 1/3 of the distance between two equally charged positive
objects. The force on the test charge exerted by the near object is F. Determine the force
exerted by the the further object on the test charge, and hence the resultant force on the test
charge.
Force on q+ from q1 = F towards q2
Force on q+ from q2 = F/4 towards q1
Resultant force on q+ is 3/4×F towards q2
What is the new resultant force if the charge on the further object is:
(a)
made 3 times as great
Force on q+ from q2×3 = 3×F/4 towards q1
Resultant force on q+ is F/4 towards q2
(b)
made 3 times as great and negative?
Force on q+ from q2×-3 = -3×F/4 towards q1 = 3×F/4 towards q2
Resultant force on q+ is 7/4×F towards q2
1
4.
You are supplied with the formula:
q 1q
d2
F = k
2
where k has a value of 9×109 Nm2C-2 in a vacuum.
(a)
What does each symbol in the formula represent, and what is the situation that the formula
describes?
Book work.
(b)
What is the force between a positive charge of 5 microcoulombs and a negative charge of
8 microcoulombs separated in air by a distance of 0.4m?
+ q 1 = 5 × 10
−6
coulombs
− q
−6
coulombs
2
= 8 × 10
d = 0.4m
−6
q 1q 2
× 8 × 10 − 6
9 5 × 10
F=k
= 9 × 10
d2
(0 . 4 )2
= 2.25 Newtons attraction
(c)
Two equal positive charges experience a force of repulsion of 3.6N when placed 30cm apart
in air. What is the value of each charge?
q1 = q
2
F = 3.6N
d = 0.3m
F = k
q 1q
d 2
3 . 6 = 9 × 10
q = 6 × 10
5.
2
9
-6
q 2
(0 . 3 )2
coulombs
How far apart in air must two charges be placed, each having a net charge of 2 coulombs, so
that the force between them is 1 newton?
q1 = q
2
= 2 coulombs
F = 1 Newton
d = ?
F = k
q 1q
d 2
2
2 × 2
d 2
d = 1 . 9 × 10 5 metres
1 = 9 × 10
9
2
6.
Two equal, small conducting spheres carrying charges of +1.0×10-7C and –2.4×10-7C are
touched together and then placed 3.5cm apart in air. With what force do they act on each
other? Is the force one of attraction or repulsion?
q 1 = + 1 . 0 × 10
q
2
= − 2.4 × 10
−7
-7
coulombs
coulombs
When charges are touched they share charge and when they are separated have half of the
resultant charge.
Net charge is -1.4×10-7 coulombs,
so q1=q2=-0.7×10-7 coulombs
d=3.5cm=0.035m
F = k
q 1q
d 2
= 9 × 10
2
= 3 . 6 × 10
7.
-2
0.7 × 0 . 7 × 10
(0.035 )2
9
Newtons
− 14
repulsion
Four identical conducting spheres, A, B, C and D are suspended by silk threads. Ball A is
touched by a charged rod and then each of the other three is brought into contact with A and
removed, one at a time. When A and B are placed with their centres 5cm apart, they repel
each other with a force of 1.6×10-6N. How much charges was initially given to A?
Consider A to have an initial charge of q. So when A & B are touched and separated, A has q/2
and B has q/2.
So when A and C are touched and separated, A has q/4 and C has q/4.
So when A and D are touched and separated, A has q/8 and D has q/8.
When A (q/8) and B (q/2) are placed 5cm apart F=1.6×10-6N.
q
F = k
A
d
q
B
2
q
1 . 6 × 10
− 6
= 9 × 10
9
q = 2 . 7 × 10
8.
× q
8
2
(0.05 )2
-9
coulombs
The electron in a hydrogen atom can be considered to be approximately 5.0×10-11m away
from the proton nucleus. What is the force of attraction between the proton and the electron?
d = 5 . 0 × 10
− 11
m
q 1 = − 1 . 6 × 10
-19
coulombs
(electron)
= + 1 . 6 × 10
-19
coulombs
(proton)
q
2
q q
F = k 1 2 2 = 9 × 10
d
= 9 . 2 × 10
−8
9
Newtons
− 1 . 6 × 10
− 19
(5 × 10
× 1 . 6 × 10
− 11
− 19
)
2
attraction
3
9.
(a)
Find the strength and direction of the electric field at a point 10cm due north of a small shere
carrying a charge of 3×10-6C. What would be the force on an electron placed at that point?
−6
q
9 3 × 10
E = k 2 = 9 × 10
d
(0.1 )2
= 2 . 7 × 10 6 Newtons/co ulomb North (0 ° T)
If q = -1.6 × 10
(b)
coulombs.
F = Eq = 2.7 × 10
= 4 . 3 × 10
10.
-19
− 13
6
× 1.6 × 10
Newtons
- 19
South
(180 ° T)
Copy the sketches shown below and then complete each sketch by drawing in electric lines of
force. Use an arrow to show which way an electron would move if placed at P.
-
+
+
+
+
+
T he a rro w s re p re se nt the d ire ct io n o f the E fie ld .
A n e le c tro n w ill tra ve l in the o p p o site d ire c t io n.
11.
An uncharged metal sphere is suspended between two plates as shown below. Which of the
fields sketched is a correct representation of the resultant field?
A
+
B
-
D
C
+
-
+
-
+
-
E
+
(d)
-
since dipole interaction will give hollow sphere a charge distribution relative to imposed
electric field.
4
12.
Below is a representation of the electric field between two metal places.
P
+
(a)
-
What features of the lines of force indicate that the field between the metal places is uniform?
Lines are parallel and equidistant.
(b)
Which of the graphs below could represent the force on a positive charge as it moves from
point P towards plate A? Explain your answer.
B
force
force
A
d is ta nc e fro m P
d is ta nc e fro m P
force
C
d is ta nc e fro m P
b - force is constant.
13.
A positive charge of 5×10-5C experiences a force of 0.3N in an electric field. What is the
electric field strength at that point?
q = + 5 × 10
-5
coulombs.
F = 0 .3 N
F
0 .3
E =
=
q
5 × 10
−5
= 6 × 10
3
N/C
in the
direction
of the
force
5
14.
An electron with charge 1.6×10-19 coulomb is placed in a uniform electric field of strength
100 newtons per coulomb.
q 1 = − 1 . 6 × 10
-19
coulombs
(electron)
E = 100 N/C
(a)
Calculate the force on the electron.
F = Eq = 1.6 × 10
(b)
-17
Newtons
What would be the force on a proton placed in the same field? Explain the differences in the
consequent motion of the electron and the proton.
F = 1.6
× 10
-17
Newtons
Same force but opposite direction. The acceleration would be less for a proton since it has a
greater mass.
(c)
The mass of the electron is 9.1×10-31kg. Calculate its acceleration in the electric field.
F
1.6 × 10 -17
a =
=
= 1 . 8 × 10 13 ms
− 31
m
9 . 1 × 10
(d)
−2
What would be the acceleration of an electron in a gravitational field of 9.8N.kg-1?
g=9.8ms-2
15.
Check the equation used to evaluate at any point the electric field strength due to a point
charge q and then calculated the electric field strength at a point 6cm away from a positive
point charges of 8×10-6 coulomb.
q
E = k 2 = 9 × 10
d
= 2 × 10
16.
7
NC
−1
9
8 × 10 -6
(0.06 )2
away
from the positive
charge
The electric field strength at a point 4.0cm from an isolated point charge is 6.0NC-1. What is
the electric field strength at a point 2.0cm from the charge?
(d 1 )
1
E2
=
E ∝ 2 ⇒
d
E1
(d 2 )2
2
42
E 2 = 6 2 = 24NC
2
−1
6
17.
Two small charged spheres carrying charges of +1.0×10-8C and +4.0×10-8C respectively are
placed 6cm apart. Find the location of the point between them at which the electric field
strength is zero.
E at P = 0
= E at P due to q 1 + E at P due to q
0 = k
q1
q
− k
2
d
d
2
2
2
Let distance between q1 and P be x, then distance between P and q2 is 6-x.
0 = 9 × 10
1
=
x2
9
1 × 10
x2
-8
- 9 × 10
9
4 × 10 -8
(6 - x )2
4
(6 − x )2
4 x 2 = (6 − x ) = 36 - 12 x + x 2
2
3 x 2 + 12 x − 36 = 0
(3 x + 18 )( x − 2 ) = 0
so
(x
- 2)= 0
x = 2cm
18.
Two point charges are placed 10cm apart in air as shown below. One is a positive charge of
8×10-6 coulomb, the other a negative charge of 1.6×10-5 coulomb. Calculate the resultant
electric field strength at the points:
(a)
M which is midway between the two charges, and
E at M = E at M due to q
= k
qA
d2
= 9 × 10
(b)
(towards
9
8 × 10
(0 . 05
= 2 . 88 × 10
7
= 8 . 64 × 10
7
A
qB
(towards
d2
1 . 6 × 10 − 5
+ 9 × 10 9
(0 . 05 )2
q
−6
)2
+ E at M due to q
B
)+
+ 5 . 75 × 10
NC
−1
k
B
q
B
)
7
towards
the negative
charge
P which is 6cm from the positive charge and 8cm from the negative charge.
E at P = E at P due to q
q
= k A2 (towards
d
= 2 × 10 7 (towards
= 3 . 0 × 1 0 NC
7
−1
A
+ E at P due to q
B
q
NE ) + k B2 (towards
SE )
d
NE ) + 2 . 25 × 10 7 (towards
in a direction
SE
)
41.5 ° to PA
7
19.
Determine the resultant electric field strength at the centre of an equilateral triangle if three
point charges of the same magnitude and sign are situated at the vertices. What would be the
field strength if one of the charges were opposite in sign to the others?
E due to q1 + E due to q2 + E due to q3 = 0
If q2 is negative the E will be in opposite direction to E from +q2. Magnitude will be double
that of q2 alone.
20.
Charged latex spheres which would normally fall under the action of gravity are held
stationary in the electric field between two charged plates. The electric field is adjusted until
the electric force on a sphere just balances the force downwards due to the gravitational field.
If a sphere of mass 2.4×10-12kg is held stationary by an electric field of 4.9×104 newtons per
coulomb, calculate the charge on the latex sphere. How many excess electrons are there on the
sphere?
m = 2.4 × 10
-12
E = 4.9 × 10
F up = Eq
balanced
× 10
4.9
= 4.8
ne
n =
21.
× 10
4.8
1.6
- 16
× 10
× 10
× 10
= mg
Eq
× q = 2.4
4
q = 4.8
-1
NC
= mg
F down
When
4
kg
× 10
- 12
× 9 .8
coulombs
- 16
- 16
− 19
= 3000
An alpha particle is a He nucleus; it has a mass of 6.68×10-27kg and a net positive charge of
2×1.6×10-19C. What strength and direction of electric field is required to support its weight at
the surface of the earth?
m = 6 . 68 × 10
− 27
q = 3.2 × 10
- 19
g = 9.8Nkg
-1
kg
C
If weight
supported
then
FG = FE
mg
6.68
× 10
- 27
× 9.8
= Eq
= E × 3.2 × 10
E = 2.05
× 10
-7
- 19
NC
-1
upwards
8