Question

Question: A particle of charge $\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{\text{C}}$ is 12 cm distant from a second particle of charge$\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{}\mathbf{\text{C}}$ . Calculate the magnitude of the electrostatic force between the particles.

Short answer

Answer:

The magnitude of the electrostatic force between the particles is 2.81 N

Step-by-step solution

The given data

The charges on the first and second particle are $q_1=+3.00\times {10}^{-6}Cand{q}_2=-1.5\times {10}^{-6}C$ , respectively.

The separation between the particles, $r=12.0cm\left(\frac{1m}{100cm}\right)=0.12m$

understanding the concept of Coulomb’s law

According to Coulomb's Law of electrostatic attraction or repulsion within particles, the force acting on them is given as being directly proportional to the product of the charges on the particles and being inversely proportional to the separation between them. Using this concept, we can find out the force acting on them.

Formula:

The magnitude of the electrostatic force between any two particles,

$F=k\frac{\left|q_1\right|\left|q_2\right|}{r^2}$ (1)

Calculation of the magnitude of the force

Substituting the given data in equation (1), we can get the magnitude of the electrostatic force as given:

$$\begin{gathered} F=\frac{\left(9\times {10}^9\frac{N·m^2}{C^2}\right)\left(3.00\times {10}^{-6}C\right)\left(1.5\times {10}^{-6}C\right)}{{\left(0.12m\right)}^2} \\ =2.81N \end{gathered}$$

Hence, the value of the force is 2.81