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crimeas [40]
1 year ago
6

Point charge 3.0 μC is located at x = 0, y = 0.30 m, point charge -3.0 μC is located at x = 0 y = -0.30 m. What are (a)the magni

tude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.5 μC at x = 0.40 m, y = 0?
Physics
1 answer:
fiasKO [112]1 year ago
4 0

The total electric force that these charges exert on a third point charge is 11.9 N and the direction is towards positive y axis.

<h3>Magnitude of the electric force between the charges</h3>

The force between the first charge and the third charge is calculated as follows;

F₁₃ = kq₁q₃/r₁₃²

F₁₃ = (9 x 10⁹ x 3 x 10⁻⁶ x 4.5 x 10⁻⁶) / (0.1)²

<u>note:</u><em> distance between charge 3.0 μC and Q = 4.5 μC  = 0.4 m - 0.3 m = 0.1 m</em>

F₁₃ = 12.15 N

The force between the second charge and the third charge is calculated as follows;

F₂₃ = kq₂q₃/r₂₃²

F₂₃ = (9 x 10⁹ x -3 x 10⁻⁶ x 4.5 x 10⁻⁶) / (0.7)²

<u>note:</u><em> distance between charge -3.0 μC and Q = 4.5 μC  = 0.4 m - (-0.3 m) = 0.7 m</em>

<em />

F₂₃ = -0.248 N

<h3>Net force between the charges</h3>

F(net) = F₁₃ + F₂₃

F(net) = 12.15 - 0.248

F(net) = 11.9 N

Thus, the total electric force that these charges exert on a third point charge is 11.9 N and the direction is towards positive y axis.

Learn more about electric force here: brainly.com/question/17692887

#SPJ1

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Answer:

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Explanation:

Conceptual analysis

Hooke's law

Hooke's law establishes that the elongation (x) of a spring is directly proportional to the magnitude of force (F) applied to it, provided that said spring is not permanently deformed:

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Data

The data given is incorrect because if we apply them the answer would be illogical.

The correct data are as follows:

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Problem development

We replace data in formula 1 to calculate  K :

F₁= K*x₁

K=( F₁) / (x₁)

K=( 80) / (8) = 10 pounds/ inche

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Explanation:

In order to be able to solve this problem, you will need to know the value of water's specific heat, which is listed as

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4.18 J

+

increase by 1

∘

C



4.18 J

=

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∘

C

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2

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To increase the temperature of

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

4.18 J

+

increase by 1

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C



4.18 J

+

...

=

increase by n

∘

C

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n

×

4.18 J

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1

∘

C

increase in a

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sample of water. You'd need to provide it with

for 1 g of water



4.18 J

+

for 1 g of water



4.18 J

=

for 2 g of water



2

×

4.18 J

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∘

C

increase in the temperature of

m

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4.18 J

+

for 1 g of water



4.18 J

+

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=

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m

×

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C

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