Finding Zero Electric Field Points on the x-axis Due to Two Point Charges

AI Thread Summary
The discussion focuses on finding points along the x-axis where the electric field is zero due to two point charges, +q at x=0 and +2q at x=3d. Participants analyze possible regions on the x-axis, concluding that the zero electric field point is likely between +q and point P, as the stronger charge +2q influences the field more significantly. The quadratic equation derived from the electric field equations is being solved to find the exact location of the zero field point. The user expresses difficulty with the algebraic manipulation but receives guidance on how to rearrange the equation for easier solving. The conversation emphasizes the importance of understanding charge interactions and algebraic techniques in physics problems.
Davetty
Messages
2
Reaction score
0

Homework Statement



Two point charges +q and +2q are held along the x-axis, +q at x=0, +2q at x=3d.
i) which region(s) on the x-axis where the electric field due to the two charges is zero?
ii) Find the exact location in terms of d

+q_____P_______+2q
x=0 x=d x=3d

Homework Equations



E = k(q/r) Hint: solve the quadratic equation: K(q/x2)=K(2q/(3d-x)2

The Attempt at a Solution


For part i, the choices were: a) to the right of +2. b) between +2q and point P. c) between +q and point P. d) to the left of +q. e) both a and c. f) both b and d.

Since both charges are positive they would be repulsing each other so I believe that there would be a charge both to the left of +q and to the right of +2q so this cancels out a and d and consequently e and f. I chose c because I figured that +2q is a stronger charge so zero would be closer to +q.

for part ii, I attempted to solve the quadratic equation cancelling out K on both sides leaving: q/x2= 2q/(3d-x)2. I took the square of both side of the equation and ended up with: square root q/x=square root 2q/3d-x and now I'm stuck.

I'm really bad at working problems without numbers and am not very good at physics, so any help would be greatly appreciated!

Thanks in advance :smile:
 
Physics news on Phys.org
You're on the right track.

So you have

<br /> \frac{\sqrt{q}}{x} = \frac{\sqrt{2q}}{3d-x}<br />

and you want to solve that for x. That's difficult when the "x" terms are in the denominator. Try to think of how to get the x's into the numerator, or how to get rid of the fractions altogether.
 
That worked! Thanks a bunch! :)
 
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...
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}...
Back
Top