What is the Distance of Closest Approach Between Two Point Charges?

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The discussion focuses on calculating the distance of closest approach between two point charges using conservation of energy principles. The initial kinetic energy of the moving charge is converted entirely into potential energy at the point of closest approach. The relevant equation involves Coulomb's Law, expressed as U = K*[q*q/r], where U is potential energy, K is Coulomb's constant, and r is the distance sought. Participants confirm that the total energy remains constant throughout the process, leading to the conclusion that the closest distance of approach is approximately 0.404 m. This problem illustrates the application of energy conservation in electrostatics.
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Homework Statement



Distance of Closest Approach
Use the similarity between Coulomb's Law and the Law of Universal Gravitation to calculate the distance of closest approach between a point charge of +3.40 × 10-6 C, which starts at infinity with kinetic energy of 8.70 J, and a fixed point charge of +1.15 × 10-4 C. Assume that the moving charge is aimed straight at the fixed point charge.
 
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Just use conservation of energy. What's KE+PE at infinity? At closest approach, KE=0, since the point charge has stopped. But the total energy hasn't changed, right?
 
I don't understand. Initially total energy and ending with total potential correct? How is this solved?
 
take the two point charges as system! now all the Columbian and gravitational forces are internal forces and hence mechanical energy is conserved.

just use KEinitial + PEinitial = KEfinal + PEfinal

find PEfinal and use it to find the distance at that time
 
Okay, i knew total energy is conserved. for anyone who needs it in the future, the closest distance of approach is when all kinetic energy is converted to potential energy. the problem can be solved using this equation: U = K*[q*q/r] where U = potential energy, K = coulomb's constant (8.9875E9), the q's are the respective charges of the particles, and r = distance of closest approach.
Notice that it is just r instead of r squared. this is enegry instead of charge force.
 
Is the answer to this problem 0.404 m? Thanks
 
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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