Question

Which expression gives the force \(\mathbf{F}_{12}\) on \(q_1\) due to \(q_2\), even though other charges are present?

विकल्प

• \(\displaystyle \mathbf{F}_{12}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}^{2}}\hat{\mathbf{r}}_{12}\)

• \(\displaystyle \mathbf{F}_{12}=\frac{1}{4\pi\varepsilon_0}\frac{q_1+q_2}{r_{12}^{2}}\hat{\mathbf{r}}_{12}\)

• \(\displaystyle \mathbf{F}_{12}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r_{12}}\hat{\mathbf{r}}_{12}\)

• \(\displaystyle \mathbf{F}_{12}=\frac{1}{4\pi\varepsilon_0}\frac{q_1q_3}{r_{13}^{2}}\hat{\mathbf{r}}_{13}\)

Answer 1

The force \(\mathbf{F}_{12}\) follows Coulomb's law and depends on \(q_1\), \(q_2\), and their separation \(r_{12}\). Other charges do not change this individual force.