The potential at location A is 367 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 786 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.

Vb = 227.33 V

Explanation:

Given:-

- The potential at location A, Va = 367 V

- Particle released from rest and arrives at B with speed = vb

- The potential at location C, Vc = 786 V

Find:-

The potential at location C is 786 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.

Solution:-

- The correlation between the potentials for location A, B and C. When the particle is released from location C and arrives at location B at twice the original speed.

Vb = (4Va - Vc) / 3

Vb = (4*367 - 786) / 3

Vb = 227.33 V