Calculating change in energy of a particle

AI Thread Summary
An electron moves between parallel plates with an initial potential of 30 V and a final potential of 150 V. The change in potential energy is calculated using the formula Ep2 - Ep1 = q(Vf - Vi), resulting in -1.92 x 10^-17 J, indicating a decrease in potential energy as the electron moves to a higher potential. The book states the answer is +1.92 x 10^-17 J, which is disputed by the discussion participants who agree with the calculated negative value. The consensus is that the book may be incorrect, as the reasoning for the negative potential energy change aligns with the principles of energy conservation.
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Homework Statement



An electron moves from an initial location between parallel plates where the electric potential is Vi = 30 V and Vf = 150 V

a) Determine the change in the electrons potential energy

2. Equations
J/C = V = N/C * m

The Attempt at a Solution


a) The electron is naturally moving to a location of higher energy, therefore its potential energy decreases as its kinetic energy increases.

Potential energy change = Ep2 - Ep1 = Vfq - Viq = q(Vf-Vi) = q(120 V) = -1.92 * 10^-17 J.

The sign makes sense to me, since the electron has more potential energy at Vi than Vf.

However, the book says the answer is +1.92*10^-17 J. Can someone help me out with why I am is wrong? Thanks
 
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Well your book looks like its wrong. Your qualitative reasoning seems correct to me.
 
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Are you sure they didn't ask for the change in kinetic energy? Because otherwise I have to agree with your answer as being negative.
 
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Yes I'm sure they asked for potential energy. I guess the book is wrong. Thanks for confirming this.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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