Potential Energy of a Collection of charges

AI Thread Summary
The discussion focuses on calculating the electric potential energy (U) of three equal point charges, each 1.35 μC, positioned at the vertices of an equilateral triangle with sides of 0.350 m. The formula U = k(Qq/r) is used to determine the potential energy for each pair of charges. Participants agree that since the distances and charges are identical, calculating U for one pair and multiplying by the number of pairs is a valid approach. The calculated total potential energy is approximately 0.138 J, representing the energy needed to separate the charges to infinity. This confirms the understanding of potential energy in a system of charges.
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Homework Statement



Three equal point charges, each with charge 1.35 \mu C, are placed at the vertices of an equilateral triangle whose sides are of length 0.350 m. What is the electric potential energy U of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)

Homework Equations



U=k(Qq/r)

The Attempt at a Solution



should i just find the U for each pair and then just multiply it by the number of possible pairs since the distances and charges are all equal? my answer = .138J
 
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Yup, seems correct to me. The magnitude of the total potential energy of a system of charges is the amount of energy required to separate each of the charges to infinity and thus to "break up" each pair.
 
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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