Question

Which expression gives the force on \(q_1\) due to all other charges in a system of \(n\) charges?

Options

• \(\displaystyle \mathbf{F}_1=\frac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\frac{q_i}{r_{1i}}\hat{\mathbf{r}}_{1i}\)

• \(\displaystyle \mathbf{F}_1=\frac{q_1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_{1i}}\hat{\mathbf{r}}_{1i}\)

• \(\displaystyle \mathbf{F}_1=\frac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\frac{q_i}{r_{1i}^{2}}\hat{\mathbf{r}}_{1i}\)

• \(\displaystyle \mathbf{F}_1=\frac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\frac{q_i}{r_{1i}^{2}}\)

Answer 1

The sum runs from \(i=2\) to \(n\), covering every charge other than \(q_1\). Each contribution includes the inverse-square distance factor and its unit vector.