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A 6.75 nC charge is located 1.99 m from a 4.46 nC point charge.
(a) Find the magnitude of the electrostatic force that one charge exerts on the other.
(b) Is the force attractive or repulsive?

Respuesta :

Explanation:

Given that,

Charge 1, [tex]q_1=6.75\ nC=6.75 \times 10^{-9}\ C[/tex]

Charge 2, [tex]q_2=4.46\ nC=4.46\times 10^{-9}\ C[/tex]

The distance between charges, r = 1.99 m

To find,

The electrostatic force and its nature

Solution,

(a) The electric force between two charges is given by :

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

[tex]F=\dfrac{9\times 10^9\times 6.75\times 10^{-9}\times 4.46\times 10^{-9}}{(1.99)^2}[/tex]

[tex]F=6.84\times 10^{-8}\ N[/tex]

(b) As the magnitude of both charges is positive, then the force between charges will be repulsive.

Therefore, this is the required solution.

asuwafohjames

(a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N.

(b) The force is repulsive.

(a) To calculate the magnitude of the electrostatic force that one charge exerts on the other, we use the formula below.

Formula:

  • F = kqQ/r².................. Equation 1

Where:

  • F = Force of  between the charges
  • q = First charge
  • Q = second charge
  • r = distance between the charges
  • k = coulomb's constant.

From the question,

Given:

  • q = 6.75 nC = 6.75×10⁻⁹ C
  • Q = 4.46 nC = 4.46×10⁻⁹ C
  • r = 1.99 m
  • k = 8.98×10⁶ Nm²/C²

Substitute these  values into equation 1

  • F = (6.75×10⁻⁹)(4.46×10⁻⁹)(8.98×10⁹)/1.99²
  • F = 6.83×10⁻⁸ N

(b) The force is repulsive because both charges a the same (positive).

Hence, (a) The magnitude of the electrostatic force that one charge exerts on the other is 6.83×10⁻⁸ N (b) The force is repulsive.

Learn more about electrostatic force here: https://brainly.com/question/8424563

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