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Sagot :
Answer:
The Force between the two charges is an attractive force of 16,000N
Explanation:
Expression for the electric force between the two charges is given by
F = (k*q1*q2) / r^2
Here, k = constant = 9 x 10^9 N*m^2 / C^2
q1 = - 2.0x10^-4C
q2 = + 8.0x10^-4C
r = 0.30 m
Substitute the given values in the above expression -
One charge is + and the other is a -, therefore the net force is an attractive force (opposites atract)
The attraction force is:
F= 9.0x10^9 * 2.0x10^-4 *8.0x10^-4 N/ 0.30^2
F= 16,000N
The force between the two charges will be 16,000 N.It is the force exerted by one charge on another charge.
What is electric force?
The electric force between the two charges is directly propotional to the product of the charge and inversly propotional to the square of the distance between them.
The given data in the problem is;
q₁ is the megnitude of charge 1= - 2.0x10⁻⁴ C
q₂ is the megnitude of charge 2 = + 8.0x10⁻⁴ C
k is the propotionallity constant = 9 x 10⁹ Nm² / C²
r is the seperated distance = 0.30 m
F is the electric force=?
The electric force is found as;
[tex]\rm F = \frac{K q_14q_2}{r^2} \\\\ \rm F = \frac{9 \times 10^9 \times 2.0 \times 10^{-4}}{(0.30)^2} \\\\ \rm F = 16,000N[/tex]
Hence the force between the two charges will be 16,000 N.
To learn more about the electric force refer to the link;
https://brainly.com/question/1076352
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