Question

The potential at location A is 362 V. A positively charged particle is released there from rest and arrives at location B with a speed v_B. The potential at location C is 795 V, and when released from rest from this spot, the particle arrives at B with twice the speed it previously had, or 2v_B. Find the potential at B.
V

Answer 1

increase in kinetic energy of +ve charge particle is equal to decrease in potential energy of sytem (q (Vi - Vf), where Vi and Vf are potentials at iniitial and final position of particle.
For particle going from A to B
1/2 m v_b² = q ( 362 - V_b) ...........1
vb is speed at point B and Vb is potential at point B

For particle going from C to B
1/2 m (2v_b)² = q ( 795 - V_b) ...............2

Dividing Eq. 2 by 1 we get
4 = (795 - V_b) / (362 - V_b)

1448 - 4(V_b) = 795 -V_b

3(V_b) =1448 - 795

V_b = 653 V