Magnitude of net electric field

AI Thread Summary
The discussion focuses on calculating the net electric field at the center of a square formed by four point charges, three positive and one negative, each with a magnitude of 2.2 x 10^-12 C. The initial attempt to use the formula E = (k|q|)/r^2 yielded an incorrect result, leading to confusion about whether to multiply by four. After some troubleshooting, the user successfully figured out the correct approach to combine the electric fields from the charges. The key takeaway is understanding how to properly calculate and combine the electric fields from multiple point charges in a geometric arrangement. The resolution highlights the importance of careful application of electrostatic principles in such problems.
mayo2kett
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Four point charges have the same magnitude of 2.2 10-12 C and are fixed to the corners of a square that is 3.0 cm on a side. Three of the charges are positive and one is negative. Determine the magnitude of the net electric field that exists at the center of the square.

i tried using E=(k|q|)/r^2 but it wasn't working and I'm not sure how to combine the electric fields for all of the outside charges...

-annie
 
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Why doesn't that formula work?
 
this is what i tried ((8.99e9)(2.2e-12))/((3.0e-2)^2) and i got 21.976... so now am i supposed to multiply by 4? cause that answer i got isn't right
-annie
 
Last edited:
ahha i figured it out... :)
 
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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