Electric Potential at Origin of Point Charges?

AI Thread Summary
The discussion centers on calculating the electric potential at the origin due to two equal point charges of 4.0 μC located at x1=-2.0m and x2=2.0m. Participants clarify that electric potentials from point charges add together as scalars rather than canceling each other out. Since both charges are positive, their contributions to the potential at the origin will sum, resulting in a non-zero value. The concept that voltage represents potential energy per charge is emphasized, reinforcing that energies can be positive or negative but still add up. Therefore, the electric potential at the origin is not zero.
freeofwork
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Homework Statement



Two equal point charges of magnitude 4.0 μC are situated along the x-axis at x1=-2.0m and x2= 2.0m. What is the electric potential at the origin of the xy coordinate system?


Homework Equations



ΔV= Kc (q/r)

The Attempt at a Solution



Do voltages cancel each other out at the mid point or do they add up?
 
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hi freeofwork! :smile:
freeofwork said:
Do voltages cancel each other out at the mid point or do they add up?

(can't they do both? :wink:)

voltage is (electric) potential energy per charge, and so can be positive or negative

energies add (as scalars), so voltages do also :smile:
 
so its not zero right?
 
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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