Respuesta :

Answer:

Force between two charges is given as 1.296 N

Explanation:

As we know that the two charges will attract or repel each other when they are placed near to each other

Here we know that electrostatic force is given as

[tex]F = \frac{kq_1 q_2}{r^2}[/tex]

here we have

[tex]q_1 = q_2 = 6 \times 10^{-6} C[/tex]

distance between two charges is given as

r = 0.50 m

so we have

[tex]F = \frac{(9\times 10^9)(6 \times 10^{-6})^2}{0.50^2}[/tex]

[tex]F = 1.296 N[/tex]

asuwafohjames

The force of electrical attraction is 1.295 N.

To calculate the force of electrical attraction between both positive charges, we use the formula below.

⇒ Formula:

  • F = kqq'/r²................ Equation 1

⇒ Where:

  • F = Force of attraction on the charges
  • q and q' = the two positive charges respectively
  • r = distance between the two charges
  • k = coulomb's constant.

From the question,

⇒ Given:

  • q = q' = 6.0×10⁻⁶ C
  • r = 0.50 m
  • k = 8.99×10⁹ Nm²/C²

⇒ Substitute these values into 1

  • F = (6.0×10⁻⁶)(6.0×10⁻⁶)(8.99×10⁹)/0.5²
  • F = 1.295 N

Hence, the force of electrical attraction is 1.295 N

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