Question

A positive point charge of magnitude \(q_{1}\) lies at the origin. Derive an expression for the incremental work done in moving a second point charge \(q_{2}\) through a distance \(d x\) from the starting position \((x, y, z)\), in the direction of \(-\mathbf{a}_{x}\)

Short answer

The expression for the incremental work done is given by: $$ d W = -k \frac{q_{1}q_{2}}{r^3} x d x $$

Step-by-step solution

Determine the force between the point charges

Using Coulomb's Law, we can determine the force exerted by the point charge \(q_{1}\) on the point charge \(q_{2}\). The magnitude of this force is given by: $$ F = k \frac{q_{1}q_{2}}{r^2} $$ where \(k\) is Coulomb's constant, and \(r\) is the distance between the charges, which can be found using the starting position, \((x, y, z)\). The distance is given by: $$ r = \sqrt{x^2 + y^2 + z^2} $$

Find the force vector

Next, we need to find the force vector acting on the charge \(q_{2}\). Since the force is acting along the line joining the two charges, we can find the unit vector along the position vector from the origin to the position \((x, y, z)\). The position vector, \(\mathbf{r}\), is given by: $$ \mathbf{r} = x\mathbf{a}_{x} + y\mathbf{a}_{y} + z\mathbf{a}_{z} $$ The unit vector along the position vector is then given by: $$ \mathbf{\hat{r}} = \frac{\mathbf{r}}{r} = \frac{x\mathbf{a}_{x} + y\mathbf{a}_{y} + z\mathbf{a}_{z}}{\sqrt{x^2 + y^2 + z^2}} $$ Now, the force vector \(\mathbf{F}\) acting on \(q_{2}\) is given by: $$ \mathbf{F} = F \mathbf{\hat{r}} $$

Find the work done by the force

The work done by the force \(\mathbf{F}\) when the charge \(q_{2}\) is moved through a distance \(d x\) in the direction of \(-\mathbf{a}_{x}\) can be found using the dot product of the force vector and the displacement vector. The displacement vector \(\mathbf{d r}\) is given by: $$ \mathbf{d r} = -d x\,\mathbf{a}_{x} $$ Now, the incremental work done, \(d W\), is given by: $$ d W = \mathbf{F} \cdot \mathbf{d r} $$

Calculate the incremental work done

Substitute the expressions for the force vector and displacement vector to find the incremental work done: $$ d W = (F \mathbf{\hat{r}}) \cdot (-d x\,\mathbf{a}_{x}) = (k \frac{q_{1}q_{2}}{r^2} \cdot \frac{x\mathbf{a}_{x} + y\mathbf{a}_{y} + z\mathbf{a}_{z}}{\sqrt{x^2 + y^2 + z^2}}) \cdot (-d x\,\mathbf{a}_{x}) $$ Since the dot product is only non-zero when the corresponding unit vectors match, we can simplify the expression to: $$ d W = -k \frac{q_{1}q_{2}}{r^3} x d x $$ This is the expression for the incremental work done in moving the second point charge through a distance \(d x\) in the direction of \(-\mathbf{a}_{x}\) from the starting position \((x, y, z)\).