Electron Force and Electric Field

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
3 replies · 2K views
Paul2011
Messages
9
Reaction score
0

Homework Statement



A) Three conducting spheres, a, B , and c, form
an equilateral triangle of side length "l" and have
initial charges of -2q, -4q, and 8q, respectively.
Show that the magnitude of the electrostatic force
between spheres a and c is given by

F_ac= (4q^2)/(π€_o l^2 )

B) The following steps are then taken: a and b are
connected by a thin wire and then disconnected;
b is grounded by the wire and the wire is then
removed; b and c are connected by the wire and
then disconnected. Show that the magnitudes of
the electrostatic forces between spheres a and c
and between spheres b and c, respectively, are
given by

F_ac=(3q^2)/(π€_o l^2 )
F_bc=(4q^2)/(π€_o l^2 )

Homework Equations



I think I got part A right, but part B i seem to be stuck unfortunately.
Any help would be appreciated, thanks.



The Attempt at a Solution



A)

F_ac=(-2q)(8q)/(4π€_o l^2)
F_ac=(4q^2)/(π€_o l^2)
 
Physics news on Phys.org
So, in part B, what I'm assuming is happening is that as the spheres are being connected and disconnected, charge is being transferred so that the charge of both of the spheres that are connected are the same. So as A and B are connected, their charges both become -3q. As B is grounded, it loses all its charge, but then is connected to C, which evens out so that B and C have a charge of 4q. The end result is that the charge on A is -3q, the charge on B is 4q and the charge on C is 4q.

Now do what you did last problem for the force and you'll have your answers!
 
Ok, so this is what I've done based on your insight (thanks btw).

(a+b)/2 = (-2q-4q)/2 = -3q

a=-3q , b=-3q, c=8q

a=-3q, b=0q, c=8q

(c+b)/2 = (8q+0q)/2 = 4q

a=-3q , b=4q, c=4q

F_ac = (-3q)(4q)/4π€_o l^2
F_ac = 3q^2/π€_o l^2

F_bc = (4q)(4q)/4π€_o l^2
F_bc = 4q^2/π€_o l^2

looks right to me, thanks again. I hope its right at least..
 
Yep, you got it. No problem.