91Ó°ÊÓ

First, we need to find the force acting on the electron due to the electric field \(\overrightarrow{\mathbf{E}}\). We know that \(q = -1.6 \times 10^{-19}\ \mathrm{C}\) is the charge of the electron. Therefore, the electric force \(\overrightarrow{\mathbf{F_{E}}} = q \overrightarrow{\mathbf{E}}\), giving us: $$\overrightarrow{\mathbf{F_{E}}} = (-1.6 \times 10^{-19}\ \mathrm{C})\Big(3.0 \times 10^{4}\ \mathrm{V/m} \ \mathrm{(east)}\Big) = -4.8 \times 10^{-15}\ \mathrm{N} \ \mathrm{(east)}$$

Next, we determine the magnetic force \(\overrightarrow{\mathbf{F_{M}}}\) on the electron due to the magnetic field \(\overrightarrow{\mathbf{B}}\). We can find this using the Lorentz force equation's magnetic component: \(\overrightarrow{\mathbf{F_{M}}} = q\ \overrightarrow{\mathbf{v}} \times \overrightarrow{\mathbf{B}}\). The velocity of the electron is given as \(\overrightarrow{\mathbf{v}} = 5.0 \times 10^{6} \ \mathrm{m/s} \ \mathrm{(south)}\), and the magnetic field is \(\mathbf{B}=0.080\ \mathrm{T}\ \mathrm{(east)}\). The cross product between the velocity and magnetic field vectors is: $$\overrightarrow{\mathbf{F_{M}}} = (-1.6 \times 10^{-19}\ \mathrm{C})\Big(5.0 \times 10^{6}\ \mathrm{m/s} \ \mathrm{(south)} \times 0.080\ \mathrm{T}\ \mathrm{(east)}\Big)$$ Since the cross product of two vectors is perpendicular to both input vectors, the magnetic force will be in the vertical direction (upward or downward). To determine the direction of the force, use the right-hand rule, which will give us a downward direction. $$\overrightarrow{\mathbf{F_{M}}} = 6.4 \times 10^{-14} \ \mathrm{N} \ \mathrm{(down)}$$

Now, we can calculate the total electromagnetic force on the electron \(\overrightarrow{\mathbf{F}}\) as the combination of electric and magnetic forces. Since the electric and magnetic forces are in perpendicular directions, we need to use the Pythagorean theorem to find the magnitude of the total force: $$|\overrightarrow{\mathbf{F}}| = \sqrt{(\overrightarrow{\mathbf{F_{E}}})^2 + (\overrightarrow{\mathbf{F_{M}}})^2} = \sqrt{(-4.8 \times 10^{-15}\ \mathrm{N})^2 + (6.4 \times 10^{-14}\ \mathrm{N})^2} \approx 6.5 \times 10^{-14}\ \mathrm{N}$$ The direction of the total force can be found using the arctangent function to find the angle between the electric and magnetic forces: $$\theta = \tan^{-1}\Big(\frac{\overrightarrow{\mathbf{F_{M}}}}{\overrightarrow{\mathbf{F_{E}}}}\Big) = \tan^{-1}\Big(\frac{6.4 \times 10^{-14}\ \mathrm{N}}{-4.8 \times 10^{-15}\ \mathrm{N}}\Big) \approx -86^{\circ}$$ Since the angle is negative, the total electromagnetic force acts downward and opposite to the east direction. Hence, the electromagnetic force on the electron moving due south is approximately \(6.5 \times 10^{-14}\ \mathrm{N}\) acting downward and opposite to the east direction.