Does the Equation ΔE = qV Calculate Both Kinetic and Electrical Energy?

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The equation ΔE = qV can be used to calculate both the change in kinetic energy and electrical energy when a charged particle is subjected to a potential difference. If a particle starts from rest and experiences a potential difference of 100kV, its change in kinetic energy can be expressed as the product of its charge and the potential difference. The electrical energy does work on the particle, resulting in an increase in its kinetic energy. This relationship holds true under the assumption that only electrostatic forces are acting on the particle. Thus, the equation effectively applies to both forms of energy in this context.
grscott_2000
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If I know the charge on a particle and I know that it has initial speed of zero and I also know that it is passed through a potential difference of 100kV, can I say that its change in kinetic energy is

deltaE = charge of particle(q) x potential difference(100kV)


Or is this equation just used for finding electrical energy?
 
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Both. The electrical energy does work on the particle, meaning its kinetic energy increases. So, yeah, it can be used as both (assuming there's only an electro-static potential and no other forces involved).
 
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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