Question

What is the electric force acting between two charges of -0.0050 C and 0.0050 C that are 0.025 m apart?

Answer 1

Answer:

(A) $-3.6\times 10^8\ N$

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:

$F_e=\dfrac{kq_1q_2}{d^2}\\\\Where,k\to Coulomb's\ constant = 9\times 10^9\ N\cdot m^2/C^2$

Plug in the given values and solve for electric force, $F_e$. This gives,

$F_e=\frac{(9\times 10^9\ N\cdot m^2/C^2) (-0.0050\ C)(0.0050\ C)}{(0.025\ m)^2}\\\\F_e=\frac{-2.25\times 10^{-4}\times 10^9}{6.25\times 10^{-4}}\ N\\\\F_e=-3.6\times 10^8\ N$

Therefore, option (A) is correct. Negative sign implies that the nature of electric force is attraction.