Consider a disk of radius 10 cm and positive surface charge density +3.7 mC/m2. A particle of charge -4.5 mC and mass 75. mg accelerates under the effects of the electric field caused by the disk, from a point at a perpendicular distance from the center of the disk.
The final speed of the particle is 1.0 m/s and the work done on the particle by the electric field is -3.0 mJ.
How fast and in what direction was the particle originally moving?
To determine how fast and in what direction the particle was originally moving, we need to use the work-energy principle and the concepts of electric potential energy and kinetic energy.
The work done by the electric field on the particle is given by:
Work = Change in electric potential energy
Since the work done is -3.0 mJ (negative sign indicates work done against the electric field), we have:
-3.0 mJ = Change in electric potential energy
Now, the change in electric potential energy can be calculated using the formula:
Change in potential energy = q * (Vf - Vi)
where q is the charge of the particle, Vf is the final potential, and Vi is the initial potential. However, we can rewrite this formula as:
Change in potential energy = q * ΔV
where ΔV is the change in potential between the final and initial points.
In this case, the change in potential energy is equal to the work done by the electric field, so we have:
-3.0 mJ = q * ΔV
Substituting the values, we have:
-3.0 mJ = -4.5 mC * ΔV
Simplifying, we find:
ΔV = 0.67 V
Now that we know the change in potential, we can determine the potential at the initial position (V_i) and the potential at the final position (V_f).
Since the disk has a positive surface charge density, it creates an electric field that points radially outward. Therefore, the potential at the surface of the disk is higher than the potential at a distance from the center.
To find the potential at the initial position, we can use the formula for the potential due to a uniformly charged disk:
V_i = (1 / 4πε₀) * (σ * R) * (1 / sqrt(r_i² + R²))
where ε₀ is the electric constant (8.854 x 10^-12 C²/Nm²), σ is the surface charge density (3.7 mC/m²), R is the radius of the disk (0.1 m), and r_i is the perpendicular distance from the center of the disk to the initial position.
Similarly, to find the potential at the final position, we can use the same formula but with r_f being the perpendicular distance from the center of the disk to the final position.
Now, let's calculate the values:
Using V = W/q, we have ΔV = -3.0 mJ / -4.5 mC = 0.67 V
Let's consider r_i = x (the perpendicular distance from the center of the disk to the initial position) and r_f = 0 (since the final position is at the center of the disk). Therefore, we have:
V_i = (1 / 4πε₀) * (σ * R) * (1 / sqrt(x² + R²))
V_f = (1 / 4πε₀) * (σ * R) * (1 / sqrt(0² + R²))
Now, we can equate the potential difference ΔV to the difference in potential at the two positions:
0.67 V = [ (1 / 4πε₀) * (σ * R) * (1 / sqrt(0² + R²)) ] - [ (1 / 4πε₀) * (σ * R) * (1 / sqrt(x² + R²)) ]
Simplifying, we get:
0.67 V = (σ * R) * (1 / sqrt(R²)) - (σ * R) * (1 / sqrt(x² + R²))
0.67 V = σ - (σ * R) * (1 / sqrt(x² + R²))
Now, we can substitute the known values:
0.67 V = 3.7 mC/m² - (3.7 mC/m² * 0.1 m) * (1 / sqrt(x² + 0.1 m²))
Solving the equation for x (the distance):
x = 0.41 m
Therefore, the perpendicular distance from the center of the disk to the initial position is 0.41 m. This means that the particle was originally moving at a distance of 0.41 m from the center of the disk in a perpendicular direction. The direction of the movement will be towards the center of the disk since the particle is negatively charged and is attracted to the positive surface charge.