# Electrostatic

Discussion in 'MCAT Study Question Q&A' started by BestDoctorEver, 04.04.12.

1. SDN is made possible through member donations, sponsorships, and our volunteers. Learn about SDN's nonprofit mission.
1. ### BestDoctorEverBannedBannedAccount on Hold 5+ Year Member

Joined:
09.25.09
Messages:
2,889
Status:
Medical Student
A popsitive charge Q is held fixed at the origin. A positive charge q is on the positive x-axis and is let go (Assume no friction)

Which of the following describes the theoritical velocity of q after it is let go?

A. Its velocity increases and then decreases to zero
B. Its velocity increases and then decreases, but it will never reaches zero
C. Its velocity increases ferever but never becomes greater than a certain bound
D. Its velocity increases forever without bound

Can someone explain?

3. ### milski1K member 5+ Year Member

Joined:
12.30.09
Messages:
2,649
Location:
Where the rain grows
Status:
Pre-Medical
The charges have the same sign which means that they will repel each other. The force is inversely proportional to the distance squared - the acceleration will be smaller and small as they get apart but theoretically it never will be zero. As such, the velocity will continue to increase but at a smaller and smaller rate.

If infinite velocity was a possibility, calculus is the way to find out if there is a limit to the velocity in this case or not. MCAT is not supposed to test that. Something that is tested on MCAT is relativity, which besides anything else stipulates that mass increases with speed, making the speed of light, c, an absolute maximum for anything moving.

TL;DR: velocity increases forever because the force between the two particles is always repelling. All velocities are bound by the speed of light, and so is this one.

4. ### pfaction 5+ Year Member

Joined:
03.14.10
Messages:
2,193
Location:
WV
MDApps:
Status:
Medical Student
I'm also gonna throw in my 2 cents.

F = ma = kqq/rsq
As r increases, force exponentially falls, and so does acceleration.
But if there's some acceleration, speed is increasing. As the distance increases, acceleration drastically decreases, and so does the gain in speed.
So it looks like / initially, and then levels off.

5. ### BestDoctorEverBannedBannedAccount on Hold 5+ Year Member

Joined:
09.25.09
Messages:
2,889
Status:
Medical Student
I understand why the acceleration will decrease but I cant process in my head how the velocity will continue to increase?...Can you explain it using some kind of formula?

6. ### pfaction 5+ Year Member

Joined:
03.14.10
Messages:
2,193
Location:
WV
MDApps:
Status:
Medical Student
As long as there's some acceleration the velocity will increase. The acceleration decreases rapidly as you move away from it
F=kqq/rsq; F=ma; so you can see the relationship.
As the A is decreasing, the V is still increasing, just very slow rate.

7. ### milski1K member 5+ Year Member

Joined:
12.30.09
Messages:
2,649
Location:
Where the rain grows
Status:
Pre-Medical
v=v0+ta

As long as a>0 v will continue to increase.

8. ### BestDoctorEverBannedBannedAccount on Hold 5+ Year Member

Joined:
09.25.09
Messages:
2,889
Status:
Medical Student
Got it

9. ### milski1K member 5+ Year Member

Joined:
12.30.09
Messages:
2,649
Location:
Where the rain grows
Status:
Pre-Medical
Acceleration tells you how velocity changes. If the velocity is 0 at the beginning and the acceleration is 10 m/s^2, after 1 sec the velocity is 10 m/s. Then the acceleration becomes 9 m/s^2 - after another second, at t=2 sec, the velocity is 10+9 m/s = 19 m/s. Now the acceleration is 8 m/s^2, so at t=3 the velocity is 27 m/s and so on. As long as the acceleration is positive, the velocity will continue to increase. If you are thinking about a formula v(t) to tell you the exact velocity at time t, that's non-trivial and not in the scope of MCAT. Knowing that the velocity will continue to increase if the acceleration is positive is good enough.

10. ### BestDoctorEverBannedBannedAccount on Hold 5+ Year Member

Joined:
09.25.09
Messages:
2,889
Status:
Medical Student
Got it...It's been a long time since I do kinematics. thanks

11. ### 411309zzzz

Joined:
07.17.11
Messages:
2,436
Location:
chillville
Status:
Pre-Medical
So the charge will fill a force on it no matter how far the distance?

12. ### milski1K member 5+ Year Member

Joined:
12.30.09
Messages:
2,649
Location:
Where the rain grows
Status:
Pre-Medical
Yes, although that force becomes negligible really quickly.

13. ### 411309zzzz

Joined:
07.17.11
Messages:
2,436
Location:
chillville
Status:
Pre-Medical
If the force becomes negligible then technically is there no acceleration or does the ridiculously small force still give it some acceleration?

14. ### milski1K member 5+ Year Member

Joined:
12.30.09
Messages:
2,649
Location:
Where the rain grows
Status:
Pre-Medical
It's ridiculously small acceleration. The increase of mass as the speed approaches speed of light makes it even less relevant.

For all practical calculations you don't need to worry about the force for particles miles and miles away. Questions like that exist just to make the point that the force does not really disappear, it's only infinitely small.

15. ### TXKnightBetter Known as TXK 5+ Year Member

Joined:
10.21.10
Messages:
1,041
Location:
Aeternal Vernis
Status:
Medical Student
I am looking at it from the point of view of U(electric) where,

U=kQq/r

one of these positive charges basically is increasing its electric potential energy with respect to the other. Like in gravitational PE, it tends to zero, but does not reach it.

it will accelerate because the Felectric is present, though it decreases in magnitude with the square of r.

just a different spin to what people are posting

16. ### milski1K member 5+ Year Member

Joined:
12.30.09
Messages:
2,649
Location:
Where the rain grows
Status:
Pre-Medical
That's neat. It will actually give you a lot better upper limit for the speed. I have to admit that using c as an upper bound was a bit of a cop out.