Final answer:
To find the electron's speed when it reaches the positive plate, calculate the work done by the electric field on the electron, which is then equated to its kinetic energy, and solve for the speed.
Step-by-step explanation:
The question asked relates to the speed of an electron when it reaches the positive plate of a parallel-plate capacitor. Given that the electric field strength is 2.30×104 N/C and the spacing between the plates is 0.600 mm, we can calculate the electron's final speed using energy conservation principles. We know that the work done by the electric field on the electron will be converted entirely into the electron's kinetic energy because the electron is released from rest.
Firstly, we need to calculate the work done (W), which is the product of the electric field (E), the charge of the electron (e), and the distance (d): W = E · e · d. Since we have E = 2.30×104 N/C and d = 0.600 mm = 0.600×10-3 m, we use the charge of an electron e = -1.60×10-19 C (negative sign indicates direction opposite to the field) to find the work done.
The kinetic energy (KE) of the electron when it reaches the positive plate will be equal to the work done: KE = ½mv2 = W. From this, we solve for the speed (v) of the electron.
Thus, the electron's speed can be found using the equation:
v = √(2W/m), where m is the mass of the electron (9.11×10-31 kg).
Substituting the values and carrying out the calculation will yield the electron's speed at the positive plate.