Two pith balls, each with a mass of 1.5 g, are attached to non-conducting threads and suspended from a common hook. Each thread has a length of 15.7 cm. The balls are then given an identical charge, which causes them to separate. In equilibrium, the threads are separated by an angle of 6.5°. Calculate the charge on each pith ball, in coulombs (C).
Two pith balls, each with a mass of 1.5 g, are attached to non-conducting threads and suspended from a common hook. Each thread has a length of 15.7 cm. The balls are then given an identical charge, which causes them to separate. In equilibrium, the threads are separated by an angle of 6.5°. Calculate the charge on each pith ball, in coulombs (C).
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Please help me with this question. Is it correct.
I got 5.47 x 10^-9
Transcribed Image Text:Two pith balls, each with a mass of 1.5 g, are attached to non-conducting threads
and suspended from a common hook. Each thread has a length of 15.7 cm. The balls
are then given an identical charge, which causes them to separate. In equilibrium, the
threads are separated by an angle of 6.5°. Calculate the charge on each pith ball, in
coulombs (C).
W
![May m = 1.5g = 1.5x10³ kg
length of thread = 15.7 cm = 0.157 m
and let charge on each ball is q
Here F = Coulombic repulsion force.
distance between two charges "r"
r = lo
= (0.157) *X [ T[ X 6.5]
180
⇒x= 0·0178 m
Since the changes are
in equilibrium
we can
write
Step 2
Tco₂/2
Tcos/2
from egn
T=
27
mg-0
F =
mg-0
Tase=
1.5x10²
from ean Ⓒ
F=
0.00085 N
from coulomb's law.
Cos
we have
F=
9x10³9²
Toes/2
and
and
=
M
1
471E
mg
Thine/21
1.5×103
Sin
1.5×10²
Cos (6-3)
Tsine= F =
q
z²
Tsing = F
r= lo
X 10 =
=
65² = [TX650
x 65²
= 0.00085
F
l = 15.7 cm
q
1.5 x 10²
1.5X102
0-998
(0.0178)2
9 = √√30x1018 C =
5-47 X10 ⁹ C
Hence charge on each pith ball =
11
0.00085
0.015 x 5in (6-5)
= 0.015
→ 9² = 3x10¹7 (²
5-47X10⁹c](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe8c83a1c-edc9-46ef-a853-ef492199ed80%2F35dbb8f7-fc75-412d-af01-ecc27ea96e4f%2F7zf4s8_processed.jpeg&w=3840&q=75)
Transcribed Image Text:May m = 1.5g = 1.5x10³ kg
length of thread = 15.7 cm = 0.157 m
and let charge on each ball is q
Here F = Coulombic repulsion force.
distance between two charges "r"
r = lo
= (0.157) *X [ T[ X 6.5]
180
⇒x= 0·0178 m
Since the changes are
in equilibrium
we can
write
Step 2
Tco₂/2
Tcos/2
from egn
T=
27
mg-0
F =
mg-0
Tase=
1.5x10²
from ean Ⓒ
F=
0.00085 N
from coulomb's law.
Cos
we have
F=
9x10³9²
Toes/2
and
and
=
M
1
471E
mg
Thine/21
1.5×103
Sin
1.5×10²
Cos (6-3)
Tsine= F =
q
z²
Tsing = F
r= lo
X 10 =
=
65² = [TX650
x 65²
= 0.00085
F
l = 15.7 cm
q
1.5 x 10²
1.5X102
0-998
(0.0178)2
9 = √√30x1018 C =
5-47 X10 ⁹ C
Hence charge on each pith ball =
11
0.00085
0.015 x 5in (6-5)
= 0.015
→ 9² = 3x10¹7 (²
5-47X10⁹c
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