An electron is released from rest just above a very large, negatively charged sheet, which carries surface charge density \(\sigma=5 \mathrm{nC} / \mathrm{m}^2\). When the electron is \(0.5 \mathrm{~m}\) above the sheet its speed is most nearly (A) \(4 \times 10^8 \mathrm{~m} / \mathrm{s}\) (B) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\) (C) \(5 \times 10^7 \mathrm{~m} / \mathrm{s}\) (D) \(1 \times 10^7 \mathrm{~m} / \mathrm{s}\) (E) \(300 \mathrm{~m} / \mathrm{s}\)

The electric field produced by an infinite sheet with a surface charge density \(\sigma\) is given by:$$E = \frac{\sigma}{2\epsilon_0}$$Where \(\epsilon_0\) is the vacuum permittivity (\(8.85 × 10^{-12} \mathrm{C^2} / \mathrm{N m^2}\)). Now we can calculate the electric field produced by the charged sheet. Substitute the surface charge density (\(\sigma = 5 \times 10^{-9} \mathrm{C/m^2}\)) into the equation:$$E = \frac{5 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}}$$

Force acting on the electron due to the electric field is given by:$$F = qE$$Where \(q\) is the charge of the electron, which is \(q = -1.6 \times 10^{-19} \mathrm{C}\). Now, we can calculate the force acting on the electron. Substitute the charge of the electron and the electric field calculated in Step 1 into the equation:$$F = (-1.6 \times 10^{-19})\left(\frac{5 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}}\right)$$

Work done on the electron is given by:$$W = Fd$$Where \(d\) is the distance the electron traveled, which is \(0.5 \mathrm{m}\). Now we can calculate the work performed on the electron using the force calculated in Step 2 and the distance:$$W = \left((-1.6 \times 10^{-19})\left(\frac{5 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}}\right)\right)(0.5)$$

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. Therefore, we can write:$$\Delta K = W$$Now we can find the change in kinetic energy of the electron using the work calculated in Step 3:$$\Delta K = \left((-1.6 \times 10^{-19})\left(\frac{5 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}}\right)\right)(0.5)$$

The kinetic energy of the electron is given by:$$K = \frac{1}{2}mv^2$$Where \(m\) is the mass of an electron, which is \(9.11 \times 10^{-31} \mathrm{kg}\), and \(v\) is its speed. We have the change in kinetic energy, so we can calculate the final velocity of the electron. Rearrange the equation for kinetic energy and solve for the final velocity:$$v = \sqrt{\frac{2\Delta K}{m}}$$Substitute the mass of the electron and change in kinetic energy into the equation:$$v = \sqrt{\frac{2\left((-1.6 \times 10^{-19})\left(\frac{5 \times 10^{-9}}{2 \times 8.85 \times 10^{-12}}\right)\right)(0.5)}{9.11 \times 10^{-31}}}$$ Calculate the final value for the velocity of the electron. The answer is closest to (C) \(5 \times 10^7 \mathrm{m} / \mathrm{s}\).