The potential at location A is 377 V. A positively charged particle is released

Question

The potential at location A is 377 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 799 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.

Explanation / Answer

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. (799-PotB) = 4 * (377-PotB) 799 - PotB = 1508 - 4 * PotB 3 PotB = 1508 - 799 = 709 Potential at B = 236.3 The potential at B is 236.3/3 V

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