Question

Two 3.0 g spheres on 1.0 m-long threads repel each other after being equally charged, as...

Two 3.0 g spheres on 1.0 m-long threads repel each other after being equally charged, as shown in the figure.

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What is the charge q?

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Answer #1
Concepts and reason

The concepts used to solve this problem are component of force and coulomb’s law.

First find the net force acting on the charge in horizontal and vertical directions.

Then use the coulomb’s law to find the charge.

Fundamentals

Coulomb’s law defined that the force between the two charges is directly proportional to the product of charges and inversely Proportional Square of the distance between the charges.

The expression for the force between the charges is,

bby

Here, is the Coulomb’s constant,9. & 92
are the charges, and is the distance between the charges.

The figure below showing the component of force acting on the charges,

T cos 20
1.0 m
20° 20°
\1.0 m
B / T sin 20 i
3.0g 19 C
3.0 g
9
mg

The tension in the string can resolve in to two components. T sin (20)
, Along direction and T cos(20°)
along direction.

From the given figure,

sin(200) = BE
BC =(AB)(sin (20°))
AB

Substitute for .

BC = (1.0m) (sin (20)
= 0.342 m

The distance between the two charges is,

r=2(BC)

Here, is the distance between the two charges.

Substitute 0.342 m
for .

r=2(0.342m)
= 0.684 m

The expression for the force between the charges is,

bby

Both the charges are same.

b=b = b

Substitute for and .

aby
4

The expression for the net force on the charge in direction is,

T cos (20°) -mg = 0

Here, is the mass, is acceleration due to gravity, and T cos(20°)
is the component of tension in the string in the direction.

Rewrite the expression in terms of tension.

T cos(20°) = mg

Rearrange the above equation to get ,

T=_mg
cos(20)

The expression for the net force on the charge in direction is,

F-Tsin(20°) = 0

Here, is the force between the charges and the component of tension in the string along the direction.

Rewrite the expression in terms of F.

F = T sin (209)

Substitute ty
for and (.07)sov Bu
for .

an in(20)=

Substitute 9x10 Nm C 2
for , for 9
& 92
, 0.684 m
for , for and 9.8m/s2
for .

(10-9 kg (9.8 m/s)
(3.08) 1g
sin (20°) =
_ (9x10° Nmục?)(q)
(0.684 m)
cos(20)

Rearrange the above equation to get ,

q= 7.458x10-C
7.5x10-C

Therefore, the charge is 7.5x10-C
.

Ans:

The charge is 7.5x10-C
.

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