ahhh okay
so for 2Q I get this: k(2Q)(q)/32 + k(4Q)(2Q)/d2 = 0
and for 4Q I get this: k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 = 0
My attempted work for solving for d now is as follows (where I will be solving 'q first' and then substituting back):
Using the 4Q equation:
k(2Q)(4Q)/d2 + k(q)(4Q)/(d...
Okay so the forces on each charge are in opposite directions and on one side will be negative then, like this?:
k(2Q)(q)/32 + k(4Q)(2Q)/d2 = -[ k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2 ]
Okay, I think I know what you mean now. I really might need a refresher on basic mechanics (and everything, really):
Do I sum up the forces then, on each charge, like so?:
k(2Q)(q)/32 + k(4Q)(2Q)/d2 = k(2Q)(4Q)/d2 + k(q)(4Q)/(d - 3)2
Yes, this.
I made it that way because it was across the whole distance, but I probably should have made '(k)(2Q)(4Q)/(7.2426)2' negative, would this make more sense?
(k)(2Q)(q)/32 + (k)(4Q)(q)/(7.2426 - 3)2= -(k)(2Q)(4Q)/(7.2426)2
Apologies for bumping this late I will try to be more active now. So in the beginning I had (corrected as) this:
2Q*q*k/32 = 4Q*q*k/(d - 3)2;
(d - 3)2/9 = 2;
And then I ended up later with this polynomial:
d2 - 6d - 9 = 0;
Then used the quadratic formula:
-B +- (B2 - 4ac)1/2/2a;
where a...
I forgot the '-b' at the beginning as well...
Hopefully everything is more neat now:
-b +- (b2 - 4ac)1/2 / 2a
d = [-6 +- (62 - (4)(1)(-9))]/(2)
d = [-6 +- (36 + 36)]/2
d = [-6 +- (72)]/2
d = 33 or d = -34?
Assuming using only the positive answer again:
[k(2Q)(q)]/32 + [k(4Q)(q)]/(33 -...
Still not going well for me then..
-b +- (b2 - 4ac)1/2 / 2a
d = [6 +- (24 - (4)(1)(-9))1/2]/2
d = 6 +- (24 + 36)1/2 / 2
d = 6 +- (60)1/2/2
d = (6+- 7.745)/2
d = 6.8725 or d = -0.8725 ?
I'm going to assume use only the positive answer for d ? :
[k(2Q)(q)]/9 + [k(4Q)(q)]/(6.8725 - 3)2...
Thanks, this isn't going so well for me :) :
(d - 3)2/9 = 2;
(d2 - 6d + 9)/9 = 2
d2 - 6d + 9 = 18
d2 - 6d - 9 = 0
I actually made a mistake for the quadratic formula part too in my last post:
-b +- (b2 - 4a)1/2 / 2a
d = [6 +- (24 - 4)1/2]/2
d = [6 +- (4.472)]/2
d = 5.23 or...
Homework Statement
A charge 2Q is placed a distance 'd' from charge 4Q. A third charge 'q' is placed 3m from charge 2Q directly on the line between charges 2Q and 4Q. Find d and q such that the force between charges 2Q and 4Q is equal to 0.
Homework Equations
Coulomb's law: f = k(|q1q2|)/r2...
Homework Statement
https://dl.dropboxusercontent.com/u/88040310/powerhw3.png
1. Draw the reactance diagram.
2. Calculate the per unit impedances
3. If bms #1 is operating at 30 kV, find the current
Assume base of 100 MVA, 300 kV on the generator side.
Homework Equations
P = VI
s = p +jq...
Trying to figure out the notation is too much :( Do you only use the property in post 76 for
$$ \frac {4e^{-4s}} {s^{2} + 6s + 25} $$ ? I think it is this though:
$$ (\frac {1} {4}) (e^{-3t})sin4(t - 4)u(t - 4) $$I think I got it now. Thank you for all the help (and extreme patience) :H
Do I use a different property for that then?
I'm thinking I might have to do use -dF(s)/ds = L[tf(t)]
If so, it would be this:
$$ \frac { 4e^{-4s}(2s + 6) + (16e^{-4s})(s^{2} + 6s + 25) } {(s^{2} + 6s + 25)^{2}} $$