Respuesta :
Answer:
(A) [tex]-3.6\times 10^8\ N[/tex]
Explanation:
Given:
Charge of one particle (q₁) = -0.0050 C
Charge of another particle (q₂) = 0.0050 C
Separation between them (d) = 0.025 m
We know that, from Coulomb's law, electric force acting between two charged particles is given as:
[tex]F_e=\dfrac{kq_1q_2}{d^2}\\\\Where,k\to Coulomb's\ constant = 9\times 10^9\ N\cdot m^2/C^2[/tex]
Plug in the given values and solve for electric force, [tex]F_e[/tex]. This gives,
[tex]F_e=\frac{(9\times 10^9\ N\cdot m^2/C^2) (-0.0050\ C)(0.0050\ C)}{(0.025\ m)^2}\\\\F_e=\frac{-2.25\times 10^{-4}\times 10^9}{6.25\times 10^{-4}}\ N\\\\F_e=-3.6\times 10^8\ N[/tex]
Therefore, option (A) is correct. Negative sign implies that the nature of electric force is attraction.
