The potential at location A is 379 V. A positively charged particle is released there from rest and arrives at location B with a speed vB. The potential at location C is 789 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.
Since the particle accelerates in going from A to B, location B has a lower potential than A. When it goes from C to B, it accelerates again, so B has a lower potential than C.
The potential difference VC - VB is 4 times the potential difference VA - VC, since the particle aquires four times more kinetic energy (twice the velocity) going from C to B than it does going from A to B.
379 - VB = (1/4) (789 - VB)
3/4 VB = 181.75
VB = 242.3 Volts
To find the potential at location B, we can make use of the conservation of energy principle.
Let's assume the charge of the particle is q and its mass is m.
1. Potential at A = 379 V
2. Potential at C = 789 V
The electric potential energy at location A is given by:
U(A) = q * V(A)
Since the particle is released from rest at location A, its initial kinetic energy is zero. Therefore, the total mechanical energy at location A is equal to the electric potential energy at location A:
E(A) = U(A) = q * V(A)
Similarly, the total mechanical energy at location C is equal to the sum of the electric potential energy and kinetic energy at that location:
E(C) = U(C) + (1/2) * m * (2vB)^2 = U(C) + 2 * m * vB^2
Using the conservation of energy principle, the total mechanical energy is conserved between locations A and C:
E(A) = E(C)
q * V(A) = U(C) + 2 * m * vB^2
Since the potential at B is what we are interested in, we can solve for vB in terms of the potential at B:
2 * m * vB^2 = q * V(A) - U(C)
Dividing both sides by 2m:
vB^2 = (q * V(A) - U(C)) / (2m)
Taking the square root of both sides:
vB = sqrt((q * V(A) - U(C)) / (2m))
Substituting the values of V(A), U(C), and 2vB:
vB = sqrt((q * 379 - q * 789) / (2m))
Simplifying further:
vB = sqrt((-410q) / (2m))
Therefore, the potential at B can be expressed as:
Potential at B = V(B) = vB^2 / q
Substituting the value of vB:
V(B) = ((-410q) / (2m)) / q
Simplifying:
V(B) = -410 / (2m)
Therefore, the potential at B is -410 / (2m).
To find the potential at point B, we need to use the conservation of energy principle. The change in potential energy (ΔPE) of the particle as it moves from one point to another is equal to the change in kinetic energy (ΔKE) of the particle.
Let's denote the potential at point A as VA and the potential at point C as VC. We are given that the potential at point A is 379 V and the potential at point C is 789 V.
The initial potential energy at point A is given by the formula PE_A = q * VA, where q is the charge of the particle. Since the charge of the particle is not given, we can conclude that it cancels out in the calculations and we can ignore it.
The initial kinetic energy at point A is zero, as the particle is released from rest.
Therefore, the initial total mechanical energy at point A, E_A, is:
E_A = PE_A + KE_A = 0 + 0 = 0
When the particle reaches point B, the potential energy at point B, PE_B, is given by PE_B = q * VB, and the kinetic energy at point B, KE_B, is given by KE_B = (1/2) * m * vB^2, where m is the mass of the particle.
The total mechanical energy at point B, E_B, is:
E_B = PE_B + KE_B = q * VB + (1/2) * m * vB^2
Similarly, when the particle reaches point C, the potential energy at point C, PE_C, is given by PE_C = q * VC, and the kinetic energy at point C, KE_C, is given by KE_C = (1/2) * m * (2vB)^2 = 2 * (1/2) * m * vB^2 = m * vB^2.
The total mechanical energy at point C, E_C, is:
E_C = PE_C + KE_C = q * VC + m * vB^2
According to the conservation of energy principle, the total mechanical energy is conserved. Therefore, E_A = E_C, or 0 = q * VC + m * vB^2. Rearranging this equation, we get:
VB^2 = -q * VC / m
Finally, we can substitute this expression for VB^2 into the equation for E_B to find the potential at point B, VB:
E_B = q * VB + (1/2) * m * vB^2
0 = q * VB - (1/2) * q * VC
q * VB = (1/2) * q * VC
VB = (1/2) * VC
Therefore, the potential at point B is half the potential at point C.
In this case, the potential at point B is:
VB = (1/2) * 789 V
VB = 394.5 V
So, the potential at point B is 394.5 V.