Middle School

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An object with a charge of [tex]4.3 \times 10^{-5} \, \text{C}[/tex] pushes another object 0.31 micrometers away with a force of 7 N. What is the total charge of the second object?

Answer :

Answer:

Charge on the other particle is given as

[tex]q_2 = 1.74 \times 10^{-18} C[/tex]

Explanation:

As we know that the force between two charges is given as

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we know that

[tex]F = 7 N[/tex]

[tex]r = 0.31 \mu m[/tex]

[tex]q_1 = 4.3 \times 10^{-5} C[/tex]

now we have

[tex]7 = \frac{9 \times 10^9 (4.3 \times 10^{-5}q_2}{(0.31\times 10^{-6})^2}[/tex]

now we have

[tex]q_2 = 1.74 \times 10^{-18} C[/tex]

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Rewritten by : Barada

Final answer:

The total charge of the second object is 0.00028337 C.

Explanation:

In order to find the total charge of the second object, we need to use Coulomb's law. Coulomb's law states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

We can use the formula F = k * (q1 * q2) / r², where F is the force, k is Coulomb's constant, q1 and q2 are the charges of the objects, and r is the distance between them.

In this case, we can rearrange the formula to solve for q2, the charge of the second object. q2 = (F * r²) / (k * q1)

Plugging in the given values, q2 = (7 N * (0.31 x 10⁻⁶ m)²) / (9 x 10^9 N m²/C² * 4.3 x 10⁻⁵ C) = 0.00028337 C

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