An electron is projected with an initial speed of (1.6x10^6 m/s) into the uniform field between two parallel plates. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates
a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
b) Suppose that the electron is replaced by a proton with the same initial speed. Would the proton hit one of the plates? If the proton does not hit one of the plates, what is the magnitude and direction of its vertical displacement as it exits the region between the plates.
c) Compare the paths traveled by the electron and the proton and explain the differences.
I am really confused
I don't know help me
Well I did a similar problem but the speed was 1.5 instead of 1.6 and the answer there is 320 N/C, the proton doesnt hit the plates, the vertical displacement is 2.73*10^-6m and the displacement is downward.
15 m
364N/C
Please need the steps to solve.
Thanks.
Please I need the steps to solve this problem.
To find the answers to these questions, we can make use of the principles of electric fields and motion. Let's break down each question and explain how to arrive at the answers:
a) To find the magnitude of the electric field, we need to consider the forces acting on the electron. When the electron enters the electric field, it experiences a force due to the electric field strength. This force is given by the equation F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength.
Since the electron has negative charge, it experiences a force in the opposite direction to the electric field. So, in order for the electron to just miss the upper plate, the electric field strength needs to be equal to the gravitational force on the electron, mg.
The mass of an electron is very small compared to the mass of a proton, so we can neglect the gravitational force. Therefore, we equate the electric force on the electron to zero to find the electric field strength:
qE = 0
(-e)(E) = 0
E = 0
Hence, the magnitude of the electric field is zero.
b) The force on a charged particle moving in an electric field is given by F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength.
If we consider a proton with the same initial speed as the electron, the force on the proton would be different because its charge is positive. Since the electric field is directed vertically downward, the force on the proton would be in the same direction as the electric field.
Due to the force acting in the same direction as the proton's initial motion, the proton would accelerate in the downward direction as it enters the electric field. This means that the proton would hit the lower plate.
c) The paths traveled by the electron and proton would be different due to their opposite charges and the direction of the electric field.
The electron, being negatively charged, experiences an upward force due to the electric field strength and thus curves upwards. It follows a parabolic trajectory and misses the upper plate.
On the other hand, the proton, being positively charged, experiences a downward force due to the electric field strength and thus curves downwards. It follows a parabolic trajectory and hits the lower plate.
The differences in charge and the direction of the electric field result in the electron and proton following different paths within the electric field region between the plates.