Electric Field and the position of a charge

AI Thread Summary
A -10.0nC charge is located at (2.0cm, 1.0cm), and the electric field is given as -225,000i N/C. The calculation for the distance from the charge to the point where the electric field is measured is correct at 2cm, but the location must be along the x-axis. Since the electric field from a negative charge points inward, the correct position should be on the line parallel to the x-axis through the charge's location. The initial answer was incorrect because it was placed on the y-axis instead of the x-axis. Understanding the direction of the electric field is crucial for determining the correct position.
dalson
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Homework Statement



A -10.0nC charge is located at position (x, y) = (2.0cm, 1.0cm) . At what (x, y) position(s) is the electric field - 225,000i N/C?


Homework Equations


E = (Kq/r^2)*r(hat)


The Attempt at a Solution



r = sqrt(Kq/E) = .02m = 2cm
So coordinates are (0,-2)


I tried putting this answer into mastering physics online and it says it is incorrect. I can't seem to find what I have done wrong. Any help please?
 
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Right away without calculating anything, I can tell you that since the electric field of a negative point charge points radially inward in all directions, the only way for the field to be pointing in the x-direction is if the point at which you are measuring the field is somewhere along a line that runs parallel to the x-axis and passes through the charge. Your answer cannot be right, because your point lies on the y axis.

Hint: r only tells you the distance of the point away from the charge...it does not specify in which direction.
 
Yeah, you've got the right distance but the wrong location. You've even got the right idea, using the negative sign, but like cepheid says, the field points in the i direction.
 
Ah, I get it now! Thank you very much!
 
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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