5 Electrical grids are placed 1m apart and connected to batteries so that the grids are at 0V, +100V, -10V, +45V, and +75V. A dust particle with mass of 1mg and a charge of -5e enters the first grid with a KE of 100eV.
A: Will the dust particle have enough energy to make it all the way to the last grid? If not, how much extra KE does it need?
B: With the energy from part A adjusted as necessary to allow the dust particle to make it through, what KE will it have at each of the five grids, in eV?
C: What will be the velocity of the particle in m/s when leaving the 5th grid?
To answer these questions, we need to consider the energy changes as the dust particle moves through the electrical grids. Let's break it down step by step.
A: Will the dust particle have enough energy to make it all the way to the last grid? If not, how much extra KE does it need?
To determine if the particle has enough energy, we need to calculate the total potential energy change across all the grids. The potential energy change is given by the equation: ∆PE = q∆V, where q is the charge (in coulombs) and ∆V is the potential difference (in volts).
In this case, the charge of the dust particle is -5e (negative because it is negatively charged) and the potential differences are as follows:
∆V1 = 100V (difference between the first grid and the reference point),
∆V2 = -10V,
∆V3 = 45V,
∆V4 = 75V,
∆V5 = 0V (last grid).
Let's calculate the total potential energy change:
∆PE_total = q(∆V1 + ∆V2 + ∆V3 + ∆V4 + ∆V5)
= (-5e)(100V - 10V + 45V + 75V + 0V)
= (-5e)(210V)
= -1050eV
Since the potential energy change is negative, it means the particle loses energy as it moves through the grids. So, the dust particle does not have enough energy to make it all the way to the last grid.
To calculate how much extra KE it needs, we can use the conservation of energy. The initial KE of the particle is given as 100eV. The total energy should be conserved, so the sum of the initial KE and the extra KE needed should equal the total potential energy change (which is negative).
Extra KE needed = -∆PE_total - initial KE
= -(-1050eV) - 100eV
= 1050eV - 100eV
= 950eV
Therefore, the dust particle needs an extra 950eV of kinetic energy to make it through all the grids.
B: With the energy from part A adjusted as necessary to allow the dust particle to make it through, what KE will it have at each of the five grids, in eV?
To answer this question, we need to distribute the extra KE of 950eV across the grids based on the potential differences.
Since the particle starts with 100eV of KE, the remaining 950eV will be distributed as follows:
KE1 = initial KE + extra KE1 (between the first and second grid)
= 100eV + ∆PE1 (where ∆PE1 is the potential energy change across the first grid)
= 100eV + (-5e)(100V)
= 100eV - 500eV
= -400eV (negative sign because it loses energy)
KE2 = KE1 + ∆PE2
= -400eV + (-5e)(-10V)
= -400eV + 50eV
= -350eV
KE3 = KE2 + ∆PE3
= -350eV + (-5e)(45V)
= -350eV - 225eV
= -575eV
KE4 = KE3 + ∆PE4
= -575eV + (-5e)(75V)
= -575eV - 375eV
= -950eV
KE5 = KE4 + ∆PE5
= -950eV + 0eV
= -950eV
Therefore, the KE at each of the five grids will be:
KE1 = -400eV
KE2 = -350eV
KE3 = -575eV
KE4 = -950eV
KE5 = -950eV
C: What will be the velocity of the particle in m/s when leaving the 5th grid?
To calculate the velocity of the particle, we can use the equation: KE = (1/2)mv^2, where m is the mass of the particle (1mg) and v is the velocity.
Since the mass of the particle is 1mg (0.001g = 0.000001kg), and the KE at the 5th grid is -950eV, we can convert the KE to joules and solve for v:
KE5(J) = -950eV * (1.6 x 10^-19 J/eV)
= -1.52 x 10^-16 J
KE5(J) = (1/2)mv^2
-1.52 x 10^-16 J = 0.5 * 0.000001kg * v^2
v^2 = (-1.52 x 10^-16 J) / (0.5 * 0.000001kg)
v^2 = -3.04 x 10^-10 m^2/s^2
Taking the square root, we find:
v = sqrt(-3.04 x 10^-10 m^2/s^2)
Note that the negative sign indicates the direction of charge (negative). However, taking the square root of a negative value results in an imaginary number. Therefore, the velocity cannot be calculated using this method.
In conclusion, since we cannot accurately determine the velocity using the given information, we need additional information or approach this problem from a different angle.