Recent content by Final_HB

when you say resolve perpendicular/parallel to the slope, do you mean like: F=Fcosα + Fsinα

Well... F1+F2+F3+F4+F5=0 And for the second: F1+F2-F3-F4-F5=0 taking one away from the other gives us: 2F1+2F2=0 So, F1+F2=0

Homework Statement A force of magnitude F acting up and along a smooth inclined plane, can support a mass M in equilibrium. If a force of the same magnitude acts horizontally, it can support a mass m on the same inclined plane in equilibrium. Find a relationship between F, M and m which is...

Homework Statement A particle is under the influence of 5 forces. Three of these forces are reversed, and the particle remains in equilibrium. Prove that the particle will remain in equilibrium even if these three forces were removed altogether. The Attempt at a Solution My thinking for...

Brilliant :smile: Thank you so much seriously. :smile:

So its just -2 to cancel out the +2. And since there is no extra charge going in... the charge outside the shell is -4. So to find the electric field just outside the surface. we take that charge of -4 and sub it into the formula for an electric field. (ie. kQ/r2) EDIT: just noticed the...

yes I have :smile: So there is a -2 charge induced, which then adds to the -6 that we have already... So total charge induced is -8 ??

Beautiful :smile: So the energy has decreased by 3.55x10-6

And we were just getting places :frown: am... it comes up as 1.1505x10-8 so energy stored is : 1/2CV2= 1.165x10--5 (not a massive change)

yes, just at #15. After that... nothing :frown: ya, its not pretty... but its not art class either. If there is a +2 charge enlcosed... and we know that all charge enlcosed is 0... there has to be a charge cnacelling out the +2, yes?

Sorry... Had electric field units in my head. The original energy in the system would be found the same way. And it comes up as 1.52x10-5 So energy on the plates has increased by 1.4618x10-5 Joules.

short and sweet way it is :smile: E=1/2CV2 E=\frac{1}{2} (5.7525x10-10 ) (45)2 E=5.82x10-7 V/M Right?

Im sorry... you lost me :frown: I am not sure where this is going anymore... Late, I know... Internet went down and we need a replacement.

so... we get the original charge before any change in dielectric, and find the new charge and do a little subtraction? Im REALLY sorry for being a month late to do this... my internet shut down and we need to get a replacement :frown:

Beautiful :smile: So for part b) Same story? Do i just sub into 1/2CV2 ?