A charge of 13.0 C is moved through a potential difference of 40 V over a distance of 5.0 mm. Calculate the work done and the magnitude of the electric field.

Step 1: Identify the given values.
Potential difference $\Delta V = 40 \, \text{V}$.
Charge $q = 13.0 \, \text{C}$.
Distance $d = 5.0 \, \text{mm} = 5.0 \times 10^{-3} \, \text{m}$.

a) Calculate the work done.
The work done $W$ to move a charge $q$ through a potential difference $\Delta V$ is given by:

$$W = q \Delta V$$

Substitute the given values:

$$W = (13.0 \, \text{C}) (40 \, \text{V})$$ $$W = 520 \, \text{J}$$

The work done is 520 J.

b) Calculate the magnitude of the electric field.
For parallel plates, the magnitude of the electric field $E$ is related to the potential difference $\Delta V$ and the separation $d$ by:

$$E = \frac{\Delta V}{d}$$

Substitute the given values:

$$E = \frac{40 \, \text{V}}{5.0 \times 10^{-3} \, \text{m}}$$ $$E = 8000 \, \text{V/m}$$

The magnitude of the electric field is 8000 V/m.

The final answers are:

a) $\boxed{\text{520 J}}$

b) $\boxed{\text{8000 V/m}}$