How Does Charge Polarity Affect Electrostatic Equilibrium Position?

AI Thread Summary
The discussion revolves around determining the position for a third point charge, q = -2.62 nC, where the net electrostatic force on it is zero, given two other charges, Q1 = 5.23 nC and Q2 = 1.89 nC. Initially, the charge q was successfully placed at 1.643 m when Q2 was positive. However, when the sign of Q2 is changed to negative, the user struggles to find the correct position for q, initially calculating 2.25 m but doubting its accuracy. The conversation emphasizes the importance of defining the regions where the charge can be placed and ensuring all distance variables are positive for clarity. Ultimately, the correct approach involves analyzing the regions around the charges to find a sensible solution.
badd99
Messages
21
Reaction score
0

Homework Statement



An electric point charge of Q1 = 5.23 nC is placed at the origin of the real axis. Another point charge of Q2 = 1.89 nC is placed at a position of p = 2.63 m on the real axis. At which position can a third point charge of q = -2.62 nC be placed so that the net electrostatic force on it is zero?

1.643 m (SOLVED)


Let the sign of Q2 be changed from positive to negative. At which position can the point charge q be placed now so that the net electrostatic force on it is zero?


The Attempt at a Solution



Okay so I set up k(q1)(q2)/d^2=k(q2)(q3)/(D-d)^2 where D = 2.63m

I then used the quadratic and since there is no b, I used 0.

I got:
7.12E-6d^2+0d-3.62E-5 and did the quadratic to get 2.25m but this is not right?

Help? Thanks!
 
Physics news on Phys.org
The first thing to figure out is which region the charge must be placed: to the left of Q1, between Q1 and Q2, or to the right of Q2. Only one region will give you a sensible answer.
 
so the answer should be negative then? I am still having trouble ugh.
 
badd99 said:

The Attempt at a Solution



Okay so I set up k(q1)(q2)/d^2=k(q2)(q3)/(D-d)^2 where D = 2.63m
Define d and D more clearly.

Better still, pick a region and define your distances so they are all positive.
 
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}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top