Find the work done as the particle moves along a straight line from (2, 0, 0) to (2, 1, 5).

Steps to Solve

1. Identify the force vector

The force exerted on the charged particle is given by:

$$ F(\mathbf{r}) = K \frac{\mathbf{r}}{|\mathbf{r}|^3} $$

Where $\mathbf{r} = \langle x, y, z \rangle$. To compute the force at any point, we need to calculate $|\mathbf{r}| = \sqrt{x^2 + y^2 + z^2}$.

2. Determine the path of motion

The particle moves from point A $(2, 0, 0)$ to point B $(2, 1, 5)$. We can parameterize the path using a variable $t$ where:

$$ \mathbf{r}(t) = \langle 2, t, 5t \rangle \quad \text{for} \quad t \in [0, 1] $$

This parameterization describes a straight line from A to B.

3. Calculate the force along the path

Substitute the parameterization into the force function:

$$ \mathbf{r}(t) = \langle 2, t, 5t \rangle $$

Now, compute the magnitude:

$$ |\mathbf{r}(t)| = \sqrt{2^2 + t^2 + (5t)^2} = \sqrt{4 + t^2 + 25t^2} = \sqrt{4 + 26t^2} $$

Then, the force becomes:

$$ F(t) = K \frac{\langle 2, t, 5t \rangle}{(4 + 26t^2)^{3/2}} $$

4. Find the differential displacement vector

To find the work done, we need the differential displacement vector:

$$ d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = \langle 0, 1, 5 \rangle dt $$

5. Calculate the work done

The work done is given by the integral of the force along the path:

$$ W = \int_{0}^{1} F(t) \cdot d\mathbf{r} $$

Substituting in our expressions:

$$ W = \int_{0}^{1} K \frac{\langle 2, t, 5t \rangle}{(4 + 26t^2)^{3/2}} \cdot \langle 0, 1, 5 \rangle dt $$

Calculate the dot product:

$$ W = \int_{0}^{1} K \frac{t + 25t}{(4 + 26t^2)^{3/2}} dt = \int_{0}^{1} K \frac{26t}{(4 + 26t^2)^{3/2}} dt $$

6. Evaluate the integral

To evaluate the integral, use a substitution method. Let ( u = 4 + 26t^2 ), then ( du = 52t dt ) or ( dt = \frac{du}{52t} ).

Transform the limits for $t = 0$ to $t = 1$:

When $t = 0$, ( u = 4 ) and when ( t = 1 ), ( u = 30 ).

The adjusted integral becomes:

$$ W = K \int_{4}^{30} \frac{26}{u^{3/2}} \cdot \frac{1}{52} du = \frac{K}{2} \int_{4}^{30} u^{-3/2} du $$

Evaluate the integral:

$$ W = \frac{K}{2} \left[-2 u^{-1/2}\right]_{4}^{30} = K \left[-1/\sqrt{30} + \frac{1}{2}\right] $$

7. Final expression for work done

After evaluating, the expression for work done is:

$$ W = K\left( \frac{1}{2} - \frac{1}{\sqrt{30}} \right) $$