A charged disk of radius R that carries a surface charge density σ produces an electric field at a point a perpendicular distance z from the center of the disk, given by:
Edisk = (sigma/2e0) x(1-(z/(square root(z^2 + R^2))
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?
a) 0 m/s
b) 9.0 m/s towards the disk
c) 9.0 m/s away from the disk
d) 50 km/s towards the disk
e) 50 km/s away from the disk
To determine the initial speed and direction of the particle, we can use the work-energy principle.
The work done on the particle by the electric field is given as -3.0 mJ (negative because work is done against the electric field).
The work-energy principle states that the work done on an object is equal to the change in its kinetic energy. Mathematically, this can be expressed as follows:
Work = Change in kinetic energy
The change in kinetic energy can be represented as:
ΔKE = (1/2) * m * (v^2 - u^2)
Where:
ΔKE is the change in kinetic energy,
m is the mass of the particle,
v is the final speed of the particle, and
u is the initial speed of the particle.
We are given the final speed of the particle, which is 1.0 m/s. We need to solve for the initial speed (u).
First, we need to convert the work done on the particle from mJ to Joules:
-3.0 mJ = -3.0 * 10^-3 J
Now we can substitute the values into the work-energy equation:
-3.0 * 10^-3 J = (1/2) * (75 * 10^-6 kg) * (1.0^2 - u^2)
Simplifying the equation:
-3.0 * 10^-3 J = (1/2) * (75 * 10^-6) kg * (1.0 - u^2)
Rearranging the equation:
-6.0 * 10^-3 J = 37.5 * 10^-6 kg - (37.5 * 10^-6 kg)u^2
Multiplying both sides by -1 to get rid of the negative sign:
6.0 * 10^-3 J = (37.5 * 10^-6 kg)u^2 - 37.5 * 10^-6 kg
Combining like terms:
6.0 * 10^-3 J = 37.5 * 10^-6 kg * u^2 - 37.5 * 10^-6 kg
Simplifying further:
6.0 * 10^-3 J = 37.5 * 10^-6 kg * u^2 - 37.5 * 10^-6 kg
Adding 37.5 * 10^-6 kg to both sides:
6.0 * 10^-3 J + 37.5 * 10^-6 kg = 37.5 * 10^-6 kg * u^2
Dividing both sides by 37.5 * 10^-6 kg:
(6.0 * 10^-3 J + 37.5 * 10^-6 kg) / (37.5 * 10^-6 kg) = u^2
Simplifying:
u^2 = (6.0 * 10^-3 J + 37.5 * 10^-6 kg) / (37.5 * 10^-6 kg)
u^2 = (6.0 * 10^-3 J / 37.5 * 10^-6 kg) + 1
u^2 = (160 J/kg) + 1
u^2 = 161 J/kg
Taking the square root of both sides:
u = √(161 J/kg)
Evaluating the square root:
u ≈ 12.69 m/s
Therefore, the initial speed of the particle was approximately 12.69 m/s.
Since the work done by the electric field is negative (-3.0 mJ), indicating work is done against the electric field, we can conclude that the particle was originally moving towards the disk.
Therefore, the correct answer is:
b) 9.0 m/s towards the disk