An electron moving parallel to x axis has an initial speed of 2.51x106 m/s at the origin.Its speed is reduced to 1.76x105 m/s at the point x=3.12 cm. Calculate the potential difference, in Volts, between the origin and that point.
To calculate the potential difference, we can use the formula:
ΔV = -∫E•dl
where ΔV is the potential difference, E is the electric field, and dl is the displacement vector.
First, let's calculate the electric field E at the point x=3.12 cm.
Since the electron is moving parallel to the x-axis, the electric field only has a component along the x-axis. Thus, we can write E = Exî, where Ex is the x-component of the electric field, and î is the unit vector along the x-axis.
To calculate Ex, we can use the equation:
F = qE
where F is the force experienced by the electron, q is the charge of the electron, and E is the electric field.
The force experienced by the electron is given by Newton's second law:
F = ma
where m is the mass of the electron, and a is its acceleration.
Since the electron is moving at a constant speed, its acceleration is zero. Therefore, the force experienced by the electron is also zero.
Setting F = 0 in the equation F = qE, we get:
0 = qEx
Solving for Ex, we find:
Ex = 0
This means that there is no electric field along the x-axis.
Since the electric field is zero, the potential difference ΔV between the origin and the point x=3.12 cm is also zero.
Therefore, the potential difference is 0 Volts.
To calculate the potential difference, we need to use the equation:
ΔV = -∫Edx
Where ΔV is the potential difference, E is the electric field, and dx is the displacement.
Given that the electron is moving parallel to the x-axis and its speed is reduced, we can conclude that there is a retarding force acting on the electron due to the electric field.
The electric field, E, can be calculated using the equation:
E = -dv/dx
Where v is the velocity of the electron.
To calculate E, we first need to calculate the change in velocity (dv). The change in velocity is given by:
dv = v_final - v_initial
Plugging in the values, we get:
dv = (1.76x10^5 m/s) - (2.51x10^6 m/s)
= -2.332x10^6 m/s
Given that dx = 3.12 cm = 0.0312 m, we can calculate the electric field, E:
E = -dv/dx
= (-2.332x10^6 m/s) / (0.0312 m)
= -7.48x10^7 N/C
To calculate the potential difference, we integrate the electric field over the displacement, which in this case is from x = 0 to x = 3.12 cm (or 0.0312 m).
ΔV = -∫Edx
= -∫[-7.48x10^7 N/C]dx
= 7.48x10^7 N/C * (0.0312 m - 0 m)
= 2.33x10^6 V
Therefore, the potential difference between the origin and the point x=3.12 cm is 2.33x10^6 Volts.