Solving Physics Problem: Sphere & Charge Sheet

[wormwoodsilver101]

Homework Statement

A small sphere with mass 5.00×10−7 kg and charge +6.00 μC is released from rest a distance of 0.400 m above a large horizontal insulating sheet of charge that has uniform surface charge density σ = +8.00 pC/m2.Using energy methods, calculate the speed of the sphere when it is
0.100 m above the sheet.

Homework Equations

V=integrate(E dl) from a to b
E=σ/2ε

The Attempt at a Solution

σql/2ε=0.5mv^2
(8*10^-12)(6*10^-6)(0.4-0.1)/2ε=0.5(5*10^-7)v^2
v=1.8m/s
But mastering physics keeps saying it is wrong
I don't know what did I did wrong

 

[BvU]

Hello Wormwood,

What if gravity were 19.62 m/s² instead of 9.81 m/s² ? Would it go just as fast then ?

 

[mjc123]

What about gravity?

 

[haruspex]

Both charges are positive, so why did it fall?

 

[wormwoodsilver101]

Ah, I totally ignored gravity >.<
With gravity the formula is:
mgh=σql/2ε+0.5mv^2
(5*10^-7)(9.81)(0.4-0.1)=(8*10^-12)(6*10^-6)(0.4-0.1)/2ε+0.5(5*10^-7)v^2
v=1.62m/s
The answer is finally correct.
Thanks for all the help >.<

 

[JAAZeid]

Ah, I totally ignored gravity >.<
With gravity the formula is:
mgh=σql/2ε+0.5mv^2
(5*10^-7)(9.81)(0.4-0.1)=(8*10^-12)(6*10^-6)(0.4-0.1)/2ε+0.5(5*10^-7)v^2
v=1.62m/s
The answer is finally correct.
Thanks for all the help >.<

Hey I understood the gravity was missing. But I still don't get how you arrived to this formula from the energy conservation. a and b refer to the initial and final positions.
If Ka+Uag+Uael=Kb+Ubg+Ubel and Ka=0 (at rest) and Ug=mgh and Uel=qEh
Then mgha + qEha = 0.5mv^2 + mghb + qEhb
Then v^2 = 2(mg+qE)(ha-hb)/m right? In your answer it seems like you had mg-qE instead of mg+qE!
And I think you're right about this but can you help me identify my mistake?

 

[mjc123]

What are the directions of the gravitational and electric fields? (Bear in mind F = -dE/dx)

 

[JAAZeid]

What are the directions of the gravitational and electric fields? (Bear in mind F = -dE/dx)

Ok so, both, the particle and the sheet, are positively charged therefore the electric forces are repulsive. So on the particle it's weight is mg(down) and the electric force on it is qE(up)!
But does this allow us to switch to a negative sign while working with Energies?

 

[mjc123]

Yes; gravitational potential energy increases with height, electrical decreases with height. (Actually whether the energy is positive or negative depends on what you take as your zero, which is arbitrary, but the important thing is that the change from 0.4m to 0.1m is positive for one and negative for the other.)

 

[JAAZeid]

Yes; gravitational potential energy increases with height, electrical decreases with height. (Actually whether the energy is positive or negative depends on what you take as your zero, which is arbitrary, but the important thing is that the change from 0.4m to 0.1m is positive for one and negative for the other.)

Thank you man! that was very helpful!