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9 April, 15:25

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.

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  1. 9 April, 15:45
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    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
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