Find Minimum Potential of Two Charges on x-Axis

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To find the minimum potential between two charges on the x-axis, the derivative of the potential function must be set to zero. The potential function is defined as k*q1/r - k*q2/(r-x1), where k is the Coulomb's constant, q1 is the charge at the origin, and q2 is the charge at x1 = 13.1 cm. The minimum potential occurs between the two charges, indicating a point where a test charge would experience no net force. The discussion emphasizes the importance of correctly setting up the derivative to locate this minimum point. Understanding these concepts is crucial for solving the problem effectively.
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A charge of 0.883 nC is placed at the origin. Another charge of 0.347 nC is placed at x1 = 13.1 cm on the x-axis.

At which point on the x-axis does this potential have a minimum?

I do know a few things:

A necessary condition for the potential to have a minimum is that its derivative is 0.

The minimum is between the two charges.

I need to take the derivative of:
k*q1/r - k*q2/(r-x1), but I'm not sure about how to go about that?

I'm not looking for the answer. Just need some help setting this up.
 
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There should be a point in between the two charges where a test charge experiences no force at all. Which means that you could just plonk the test charge in there. No need to do any work to get it there, you could just slip it in sideways!
 
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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