At what point along the x axis is the electric field zero?

AI Thread Summary
The discussion centers on finding the point along the x-axis where the electric field is zero due to two point charges, -9.10 μC at x = 0 cm and 21.29 μC at x = 8.00 cm. The user initially believes the zero field point is to the left of the first charge, leading to a negative x value. However, confusion arises regarding the correct formulation of the distances in the electric field equations, particularly the use of x + 8 for the second charge. Ultimately, the user is encouraged to reconsider their calculations, especially the implications of the negative x value on the distance to the second charge.
A14
Messages
3
Reaction score
0
Moved from another forum, so homework template missing
Hello, I've been trying to answering this question but my answer is always wrong.
Two point charges, -9.10 μC and 21.29 μC, are placed at x = 0 cm and x = 8.00 cm, respectively.
(a) At what point along the x-axis is the electric field zero?

What i did: i figured that the point should be on the left of the first charge, so before x=0 (should be negative).
Then i resolve E1 + E2 =0
kQ1/r^2 + kQ2/r^2 =0
(-8,19*10^4)/(x^2) + (1,916*10^5)/((x+8)^2) = 0
(8,19*10^4)/(x^2) = (1,916*10^5)/((x+8)^2)
1.916*10^5*(x^2) = 8,19*10^4 ((x+8)^2)
Then i develop and get x1= 1.51*10^1 or x2= -3.16

What am i doing wrong??

Thank you for your help!
 
Physics news on Phys.org
Why is one denominator x+8 instead of x-8?
 
A14 said:
Hello, I've been trying to answering this question but my answer is always wrong.
Two point charges, -9.10 μC and 21.29 μC, are placed at x = 0 cm and x = 8.00 cm, respectively.
(a) At what point along the x-axis is the electric field zero?

What i did: i figured that the point should be on the left of the first charge, so before x=0 (should be negative).
Then i resolve E1 + E2 =0
kQ1/r^2 + kQ2/r^2 =0
(-8,19*10^4)/(x^2) + (1,916*10^5)/((x+8)^2) = 0
(8,19*10^4)/(x^2) = (1,916*10^5)/((x+8)^2)
1.916*10^5*(x^2) = 8,19*10^4 ((x+8)^2)
Then i develop and get x1= 1.51*10^1 or x2= -3.16

What am i doing wrong??

Thank you for your help!
Is this a homework problem?
 
I don't know, for me as it is on the left of Q1, the distance to Q2 would be x (distance from point to Q1) + 8 (distance between Q1&Q2)?

Yes it is a homework problem!
 
Yes i see why it has to be negative (it has to be on the left of Q1 which is at x=0)
But -3.16 is not the right answer..
 
A14 said:
Yes i see why it has to be negative (it has to be on the left of Q1 which is at x=0)
But -3.16 is not the right answer..
Think again about that x+8. With x=-3.16, what does that make x+8?
 
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