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.
You posted this before as "Physics please clarify". See my previous answer.
Sorry I mistakenly posted the wrong question. I will post the correct question. I really appreciate all of your help! You make physics same less foreign to me. Again thanks!
To find the potential at point B, we need to use the conservation of mechanical energy for the charged particle. The mechanical energy of the particle is the sum of its kinetic energy and its electric potential energy.
The kinetic energy of the particle is given by:
K = (1/2)mv^2
Where m is the mass of the charged particle.
The electric potential energy is given by:
U = qV
Where q is the charge of the particle and V is the electric potential at a given location.
Since the particle is released from rest at location A, its initial kinetic energy is zero. Therefore, the total mechanical energy at A is given by:
E₁ = U(A) + K(A) = qVA
When the particle arrives at location B with a speed vB, the total mechanical energy at B is given by:
E₂ = U(B) + K(B) = qVB + (1/2)mvB^2
Since the particle arrives at B with twice the speed as before, the total mechanical energy at C is given by:
E₃ = U(C) + K(C) = qVC + (1/2)mq(2vB)^2
Since mechanical energy is conserved, we can equate E₁ to E₂ and E₃ to solve for the potential at B:
E₁ = E₂
qVA = qVB + (1/2)mvB^2
E₁ = E₃
qVA = qVC + (1/2)mq(2vB)^2
Now, let's simplify these equations:
qVA = qVB + (1/2)mvB^2 --(1)
qVA = qVC + 2mqB^2 --(2)
Rearrange equation (1) to solve for qVB:
qVB = qVA - (1/2)mvB^2
Substitute this expression into equation (2):
qVA - (1/2)mvB^2 = qVC + 2mqB^2
Now, solve for qVB:
qVB = qVA - (1/2)mvB^2 - 2mqB^2
Finally, substitute the given potentials at A (379 V) and C (789 V) into the equation to calculate the potential at B:
qVB = (q)(379 V) - (1/2)m(vB^2) - 2m(vB^2)
= q(379 V) - (5/2)m(vB^2)
Therefore, the potential at B depends on the charge of the particle (q), the mass of the particle (m), and the velocity at B (vB).