Where is the Zero Electric Field Point for Two Fixed Charged Particles?

AI Thread Summary
The discussion focuses on finding the zero electric field point between two fixed charged particles on the x-axis. The first particle has a positive charge, while the second has a negative charge, which means their electric fields will cancel each other out at a specific point. A participant initially calculated the zero field point at 27.85 cm but doubts its accuracy. Another contributor suggests setting the magnitudes of the electric fields equal to each other and solving for the coordinate, indicating the solution lies between -50 and -40 cm. The importance of recognizing the nature of the charges is emphasized, as the zero field point cannot be located between the two particles due to their opposing forces.
NW8800
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Hey,

I have tried this question... But I can't seem to get the answer right... I get about 27.85cm... But I think this value is incorrect...

Can someone try this, and tell me what they got/how they did it?

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.20 x 10-8 C at x = 27.5 cm and particle 2 of charge q2 = -5.66q1 at x = 54.0 cm. At what coordinate on the x-axis is the electric field produced by the particles equal to zero?



Thanks.

NW
 
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For the field to be zero, both intensities must be identical (they are of opposite sign). Write this equality. Simplify and extract the square root! You will have a simple first degree equation. I obtain a result between -50 and -40 cm.
 
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Show us your working and we can check it for you.

If one is attractive and the other repulsive that means that the coordinate cannot be between the particles. Sice your answer is, you should realize it soen't make sense.
 
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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