Use ε₀ = 8.85 × 10⁻¹² C²/(N m²). (a) A closed surface encloses a charge of +3.5 nC…

Step 1. The electric flux. Gauss's law gives $\Phi_E = \dfrac{Q}{\varepsilon_0} = \dfrac{3.5 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 395\ \text{V m}$.

Step 2. The magnetic flux. Gauss's law for magnetism says the net flux out of any closed surface is $0$ Wb. A line that leaves the surface near the other end of the magnet must come back into it. Isolated poles do not exist.

Step 3. The emf. $|\mathcal{E}| = \dfrac{\Delta\Phi_B}{\Delta t} = \dfrac{(0.040 - 0.010)\ \text{Wb}}{0.50\ \text{s}} = 0.060\ \text{V}$.