The physics of an electron moving from one plate to another.
An electron acquires 4.2 x 10-16 J of kinetic energy when it is accelerated by an electric field from Plate A to Plate B. What is the potential difference between the plates, and which plate is at a higher potential?
To determine the potential difference between the plates and which plate is at a higher potential, we need to use the equation relating kinetic energy, electric charge, and potential difference.
The equation is:
K = q * V
Where:
K is the kinetic energy acquired by the electron
q is the charge of the electron, which is 1.6 x 10^-19 C
V is the potential difference between the plates
We are given that the electron acquires a kinetic energy of 4.2 x 10^-16 J, so we can substitute this value into the equation:
4.2 x 10^-16 J = (1.6 x 10^-19 C) * V
Now we can solve for V by rearranging the equation:
V = K / q
V = (4.2 x 10^-16 J) / (1.6 x 10^-19 C)
V ≈ 26.25 volts
Therefore, the potential difference between the plates is approximately 26.25 volts. Since the electron moved from Plate A to Plate B, Plate A must be at a higher potential than Plate B.
To find the potential difference between the plates and determine which plate is at a higher potential, we can use the relationship between kinetic energy (KE) and electric potential difference (V).
The formula for the kinetic energy of a moving electron is given by:
KE = (1/2) * m * v^2
where m is the mass of the electron and v is its velocity.
Given that the electron acquires 4.2 x 10^(-16) J of kinetic energy, we can set up the equation:
(1/2) * m * v^2 = 4.2 x 10^(-16) J
It is important to note that the mass of an electron (m) is approximately 9.11 x 10^(-31) kg.
Now, we know that the kinetic energy is related to the potential energy (PE) of the electron by the equation:
PE = q * V
where q is the charge of the electron and V is the potential difference between the plates.
The charge of an electron is -1.6 x 10^(-19) C.
Also, the potential energy is equal to the difference in kinetic energy (KE) between the initial and final states of the electron:
q * V = KE_final - KE_initial
= 4.2 x 10^(-16) J - 0
Substituting the values, we have:
-1.6 x 10^(-19) C * V = 4.2 x 10^(-16) J
Solving for V:
V = (4.2 x 10^(-16) J) / (-1.6 x 10^(-19) C)
V ≈ -2.625 x 10^3 V
The negative sign indicates that the electric potential of Plate B is higher than that of Plate A. Therefore, Plate B is at a higher potential than Plate A, and the electron moves from Plate A to Plate B in the direction of the electric field.