A \(10.0-\mathrm{V}\) battery is connected to two parallel metal plates placed in a vacuum. An electron is accelerated from rest from the negative plate toward the positive plate. a) What kinetic energy does the electron have just as it reaches the positive plate? b) What is the speed of the electron just as it reaches the positive plate?

First, we need to find the potential energy of the electron when it reaches the positive plate. The potential energy gained by the electron can be calculated using the formula: Electric potential energy = Charge * Voltage We're provided with the voltage (10 V) and the charge of the electron (\(-1.6\times10^{-19}\,\mathrm{C}\)) Potential energy = \((-1.6\times10^{-19}\,\mathrm{C})(10.0\,\mathrm{V}) = -1.6\times10^{-18}\,\mathrm{J}\) This potential energy is negative since the voltage difference causes the electron to move in the opposite direction of the electric field. When it gets closer to the positive plate, the potential energy will have increased from negative to 0, as we assume it has no potential energy on the positive plate.

Now that we have the potential energy, we can find the kinetic energy by using conservation of energy. Initial energy (At negative plate) = Final energy (At positive plate) Potential energy + Kinetic energy = 0 We know the potential energy so we can now find the kinetic energy. Kinetic energy = - Potential energy = - (-1.6 x \(10^{-18}\,\mathrm{J}\)) = \(1.6 \times 10^{-18}\,\mathrm{J}\) This is the kinetic energy that the electron has just as it reaches the positive plate. #a)# Kinetic energy of the electron just as it reaches the positive plate: \(1.6\times10^{-18}\,\mathrm{J}\)

With the kinetic energy calculated, we can now find the speed of the electron right before it reaches the positive plate using the following formula: Kinetic energy = \(\frac{1}{2}\times\)m×v² Where m is the mass of the electron (9.11 × \(10^{-31}\,\mathrm{kg}\)) Rearranging this equation, we can solve for the speed (v): v = \(\sqrt{\frac{2 \times Kinetic energy}{m}}\) Plugging in the kinetic energy (\(1.6\times10^{-18}\,\mathrm{J}\)) and the mass of the electron (9.11 × \(10^{-31}\,\mathrm{kg}\)): v = \(\sqrt{\frac{2 \times 1.6 \times 10^{-18}\,\mathrm{J}}{9.11 \times 10^{-31}\,\mathrm{kg}}} = 1.87 \times 10^6\,\mathrm{m/s}\) #b)# Speed of the electron just as it reaches the positive plate: \(1.87\times10^6\,\mathrm{m/s}\)