A particle with a charge of -1.4 μC and a mass of 2.9 x 10-6 kg is released from rest at point A and accelerates toward point B, arriving there with a speed of 22 m/s. The only force acting on the particle is the electric force. What is the potential difference VB - VA between A and B? If VB is greater than VA, then give the answer as a positive number. If VB is less than VA, then give the answer as a negative number.
When the particle moved, it lost PE and converted it to KE (conservation of energy) Once it reaches point B we can calculate the KE using .5*mv^2
KE= .5*2.9*10^-6*22^2 = 7.018*10^-4
as we know that the amount of work done is equal to the gain in KE, we know that work is the potential difference times the potential difference hence
qV=KE
therefore substitute KE
-1.4*10^-6V=7.018*10^-4
we find Voltage = 501.285volts (j/c)
To find the potential difference VB - VA between points A and B, we can use the equation for the change in electric potential energy (ΔPE):
ΔPE = q * ΔV
where q is the charge of the particle and ΔV is the change in potential.
Here, the charge is -1.4 μC (microcoulombs) and the change in potential is the potential difference between A and B (VB - VA).
First, we need to find the change in electric potential energy ΔPE. We can find this using the equation:
ΔPE = PEfinal - PEinitial
where PEfinal is the electric potential energy at B and PEinitial is the electric potential energy at A.
The electric potential energy (PE) can be found using the equation:
PE = q * V
where q is the charge and V is the electric potential.
At point A, the particle is at rest, so its kinetic energy is zero. Therefore, all of its initial energy is potential energy. So, the initial electric potential energy is given by:
PEinitial = q * VA
At point B, the particle has a speed of 22 m/s, so it has kinetic energy. The final electric potential energy is given by:
PEfinal = q * VB
Now we can find the potential difference VB - VA by subtracting PEinitial from PEfinal:
ΔPE = q * VB - q * VA
Now, substituting the known values:
ΔPE = (-1.4 x 10^-6 C) * VB - (-1.4 x 10^-6 C) * VA
Simplifying:
ΔPE = -1.4 x 10^-6 C * (VB - VA)
Since the potential difference VB - VA represents the change in electric potential energy, we can also write it as ΔPE:
VB - VA = ΔPE / (-1.4 x 10^-6 C)
Now we need to find the value of ΔPE. The change in electrical potential energy can be calculated using the work-energy principle:
ΔPE = KEfinal - KEinitial
Since the particle starts from rest at point A and arrives at point B with a speed of 22 m/s, we can substitute these values into the equation:
ΔPE = (1/2) * m * (vB^2 - vA^2)
where m is the mass of the particle and vA and vB are the velocities at A and B, respectively.
Substituting the known values:
ΔPE = (1/2) * (2.9 x 10^-6 kg) * (22^2 - 0^2)
Calculating:
ΔPE = (1/2) * (2.9 x 10^-6 kg) * (484 - 0)
ΔPE = (1/2) * (2.9 x 10^-6 kg) * 484
ΔPE = 0.0007028 J
Now we can substitute this value into the previous equation to find the potential difference VB - VA:
VB - VA = (0.0007028 J) / (-1.4 x 10^-6 C)
Calculating:
VB - VA = -502 J/C
Therefore, the potential difference VB - VA between points A and B is -502 J/C.
To find the potential difference (VB - VA) between point A and point B, we need to use the formula:
VB - VA = (q * VB - q * VA) / q
Where q is the charge of the particle and is given as -1.4 μC.
But first, we need to find the work done by the electric force in moving the particle from point A to point B. We can use the work-energy theorem, which states that the work done on an object is equal to the change in its kinetic energy.
The work done by the electric force (W) can be calculated as:
W = ΔKE
Where ΔKE is the change in kinetic energy of the particle. The change in kinetic energy is equal to the final kinetic energy (KEf) minus the initial kinetic energy (KEi).
ΔKE = KEf - KEi
The initial kinetic energy is zero since the particle is released from rest.
KEi = 0
The final kinetic energy can be calculated using the formula:
KEf = (1/2) * m * v^2
Where m is the mass of the particle (2.9 x 10^-6 kg) and v is the final speed of the particle (22 m/s).
KEf = (1/2) * (2.9 x 10^-6 kg) * (22 m/s)^2
After calculating KEf, we can substitute the values back into the previous equation to find ΔKE.
ΔKE = KEf - KEi
Now we have the work done by the electric force (W). We can use the work-energy theorem to find the potential difference (VB - VA).
W = ΔKE
The potential difference (VB - VA) can be calculated by rearranging the formula as:
VB - VA = W / q
Substituting the values of W and q into the equation, we can calculate VB - VA.
VB - VA = (W) / q
Finally, calculate VB - VA to find the potential difference between point A and point B.