A small sphere with mass 5*10^-7 kg and charge is released from rest a distance of 0.500 m above a large horizontal insulating sheet of charge that has uniform surface charge density = 8*10^-12 C/m^2 .

Question

A small sphere with mass 5*10^-7 kg and charge  is released from rest a distance of 0.500 m above a large horizontal insulating sheet of charge that has uniform surface charge density = 8*10^-12 C/m^2 .

anonymous · Answer

the charge = 3*10^-6 C

jamesj · Answer

ok, so this is the set up.  What's the question?

jamesj · Answer

You want to know the net force on the sphere?  Or something else?

anonymous · Answer

i will write it again

anonymous · Answer

http://openstudy.com/study#/updates/4f54fda3e4b01ed613837ad0

jamesj · Answer

A small sphere with mass 5*10^-7 kg and charge = 3*10^-6 C is released from rest a distance of 0.500 m above a large horizontal insulating sheet of charge that has uniform surface charge density = 8*10^-12 C/m^2 

Using energy methods, calculate the speed of the sphere when it is 0.150 m above the sheet of charge?

jamesj · Answer

So what are the two forces: electric and gravity.  The change in gravitational PE I assume you know how to calculate?

jamesj · Answer

talk to me

anonymous · Answer

attachment

jamesj · Answer

Yes.  Ok.  So the KE will be the change in gravitational PE plus the change in electric PE.

Now can you calculate the first of these, the change in gravitational PE?

anonymous · Answer

what is the charge of the surface

jamesj · Answer

You are told the surface charge density, and that is what you will need to calculate the electric force, not the total charge on the surface.  If you want to go down this path now:
what is the electric field of a large charged plane with surface charge density $ \sigma $?

anonymous · Answer

can u solve the Q

jamesj · Answer

I can, but we don't give entire answers here, we help you get to them. The electric field of large charged plane with surface charge density $ \sigma $ is $$ E = \frac{\sigma}{\epsilon_0} $$ This is an important result. If you don't know it, watch this lecture: http://ocw.mit.edu/courses/physics/8-02-electricity-and-magnetism-spring-2002/video-lectures/lecture-3-electric-flux-and-gausss-law/ beginning at 23:50 Now that being the case, the electric force on the sphere is $$ F = Eq = \frac{q\sigma}{\epsilon_0} $$ where q is the charge on the sphere. As that electric field is in the vertical direction, the force is also in the vertical direction and hence the change in electric potential energy, call it EPE is $$ EPE = Fd = \frac{q\sigma d}{\epsilon_0} $$ You are given all of the variables in the numerator in your problem. The epsilon_0 term is a standard constant. So you can now calculate it.

jamesj · Answer

The change in gravitational PE you should know: it is 
$$ GPE = mgd $$
Thus the kinetic energy of the sphere is
$$ KE = EPE + GPE $$
I hope you also know of course that 
$$ KE = \frac{1}{2}mv^2 $$ and thus you can solve for speed v.

anonymous · Answer

so the final answer is .........

jamesj · Answer

You calculate it.  I've given you all the Physics now.

anonymous · Answer

what is d

jamesj · Answer

d is the distance the object moves down in the direction of gravity.