Where Along the Y-Axis is the Electric Field Zero?

AI Thread Summary
To find the point along the y-axis where the electric field is zero between two charges, the electric fields produced by each charge must be equal. The user sets the distance from the charge at the origin as R and the distance from the second charge as d-R. They derive the equation kQ/r^2 = kq/(d-r)^2 and simplify it to find a relationship between r, d, Q, and q. The user encounters difficulty in solving for r and seeks assistance with the algebraic manipulation needed to isolate r. The discussion emphasizes the importance of balancing the electric fields from both charges to determine the zero electric field point.
Punchlinegirl
Messages
221
Reaction score
0
A charge of -4.84 \mu C is located at the origin, and a charge of -6.22 \mu C is located along the y-axis at 1.361 m. The value of the Coulomb constant is 8.99 x 10^9. At what point along the y-axis is the electric field zero. Answer in units of m.
I know that for the electric field to be zero, the electric field on the charges must equal each other.
I called the distance from the charge at the origin R, and the distance from the second point is d-r.
I used the equation for electric field and set them equal to each other.
kQ/r^2= kq/(d-r)^2
Q/r^2 = q/ (d-r)^2
Q(d-r)^2 =qr^2
(d-r)^2 =qr^2/ Q
(d^2 -2dr + r^2)/ r^2 = q/Q
d^2/r^2 - 2d/r +1 = q/Q
At this point I got stuck on how to solve for r, can someone help me?
Thanks in advance.
 
Physics news on Phys.org
Here's some help
(d-r)^2 =qr^2/ Q
d-r = r \sqrt{\frac{q}{Q}}
 
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