Question

The potential at location A is 417 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.

Answer 1

Answer:

Explanation:

VA = 417 V

Vc = 799 V

Let the potential at B is V.

Mass of electron, m = 9.1 x 10^-31 kg

By the formula of energy conservation

e x ΔV = 0.5 x mv²

case I :

e x ( VB - VA) = 0.5 x m x vB²

e ( V - 417) = 0.5 mvB²     .... (1)

case II:

e x ( VB - Vc) = 0.5 x m x (2vB)²

e(V - 799) = 0.5 x 4mvB²    .... (2)

Divide equation (2) by equation (1)

$\frac{V - 799}{V - 417}= 4$

V - 799 = 4V - 1668

3V = 869

V = 289.7 V

Thus, the potential at B is 289.7 V.