Question

The potential at location A is 452 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 791 V, and when re- leased from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2vB. Find the potential at B.

Answer 1

By conservation of energy, in both cases A->B and C->B the change in potential energy has to offset the change in kinetic energy. Kinetic energy is proportional to the square of velocity.

c = Particle charge
m = Particle mass

-change in potential energy = change in kinetic energy

c * [(Potential at A)-(Potential at B)] = .5 * m(vB)^2 = K1
c * [(Potential at C) - (Potential at B)= = .5 * m (2*vB)^2 = 4 * K1

Divide through by c. The change in potential from C is 4 times that from B (the exact energy is unknown and doesn't matter). 4 times the energy is necessary to double the velocity.

(791-PotB) = 4 * (452-PotB)
791 - PotB = 1808 - 4 * PotB
3 PotB = 1808 - 791 = 1017
Potential at B = 339 V