Calculating Electric Potential and Kinetic Energy for Moving Charges

AI Thread Summary
The discussion revolves around calculating the kinetic energy and electric potential for a moving charge in an electric field. The first question involves determining the kinetic energy of a particle with a charge of 7.5x10^(-6)C as it moves from x=0.1m to x=1.0m in the presence of a fixed charge of 2.0x10^(-6)C. The user applied the relationship between kinetic energy and potential energy, leading to a calculated kinetic energy of 7.875x10^-1 J, although they seek confirmation of its accuracy. The second question addresses the electric potential at x=0.18m due to a uniformly distributed charge, with an initial calculation yielding a potential of 72, which the user doubts is correct. Overall, the user is looking for validation of their calculations and understanding of the concepts involved.
lawandb
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I have solved these and got an answer but i have no idea if i have the right work or answer. Can anyone help me with any of these questions? or just tell me what the answer is so i know if i have done it right. Thanx.

1. A particle (charge 7.5x10^(-6)C)is released from rest at a point on the x-axis, x=0.1m. It begins to move due to the presence of a 2.0x10^(-6)C charge which remains fixed at the origin. What is the kinetic energy of the particle at the instant it passes the point x=1.0m?

2. A charge of uniform density (0.80 nC/m) is distributed along the x-axis from the origin to the point x=0.1m. What is the electric potential (relative to zero at infinity) at a point x=0.18m, on the a-axis?
 
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What have you done?
 
Here's what i did

1. let *=change in
*K=-*U
*K+*U=0
*U=q(*V)
*V=V2-V1=(k(q1)/r1)-(k(q2)/r2)=(k(7.5nC))/.9-(k (2nC))/.1=1.05x10^5
q(*V)=-7.875x10^-1
*K=7.875x10^-1
--------------------------------------------------------------

2. V=k(.8nC)/.1=72, I'm pretty sure this one is wrong though.
 
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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