Sorry I got caught up in noticing my mistake with the power of ten that I didn't do it in variables.
ΔV⋅κ⋅4πε/q = [ 1/Rinital - 1/RFinal
ΔV⋅κ⋅4πε/q + 1/Rinital = 1/RFinal
Your comment + typing it out like this I realize that the R I'm solving for is the final radius not the thickness.
so...
Thanks for replying!
Okay, one thing I noticed was I accidentally use the wrong multiplier to convert the charge to coulombs, but I'm still incorrect.
20,000 = q/(4πε0κ) [1/2.25 - 1/R]
.076 = [1/2.25 - 1/R]
.368 = 1/R
R = 2.7 ...
Homework Statement
A conducting sphere has a radius of 2.25 m and carries a positive surplus charge of 35.0 mC. A protective layer of barium titanate is applied to the surface of the sphere to make it safe for laboratory workers nearby. Safety considerations dictate that the potential...
Update: even after partial integration, you end up with an exponential integral— something I’m not familiar with yet. Professor assumed we would use an integral calculator apparently. Thank you for your help, I feel like I understand the setup of these problems more than before.
Okay, so I realized that I was just evaluating without actually integrating.
∫0∞e-x/l/(x2+.04x + .022)
I’m not sure I actually know how to evaluate this integral by hand.
You are right, I didn’t even notice, haha!
For the integral, at infinity, e goes to zero and the denominator goes to infinity so I say that is 0. At 0, e-x/l becomes one and the denominator becomes .022.
If k is N*m2 / C2 we would get N/(C*m) not N/C which is what we want... so I suppose I’m...
Okay, I think I see why that's the case, but I might have to think about it more. Trying that out:
dq = q(x)dx
dq = qoe-x/l
qo = 2.9*10-6/l (from above comment)
E = k * 2.9*10-6/l ∫0∞ e-x/l / (.02 + x)2
E = k * 2.9*10-6 / (l * .022)
E = 2.3*109 N/C
Does this look correct?
I was using the value from the problem statement, 2.9 microC. I did try something else where I integrated q(x) and set it equal to 2.9μC. That resulted in an incorrect value too, but perhaps I did it wrong, I’ll type up what I did when I am no longer on mobile. Is this the correct approach...
Homework Statement
Homework Equations
dE= k dq/r2
The Attempt at a Solution
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I started off taking a derivative of q(x).
dq = -qo/l ⋅ e-x/ldx
Then, I decided that r was the distance x along the rod + .02m. r=(.02+x)
Following that, I plugged everything into the formula:
dE = k⋅qo/l...
Yeah I get that it doesn’t really need me to solve it. Got that that my previous method for work doesn’t make sense when force varies. If I solve for a with f=ma then integrate, the +C would be vo right? Then if I integrate again, the +c would be xo, which would be zero. I can solve that for t...
I see where I went wrong trying to use kinematics.
W = Δt was something I solved for
W = F*d
F = 2n/s * Δt
d = .5m
W = Δt joules?
Edit-
Just saw your edit:
I think I get it, done a bit of calculus, I was approaching it wrong from the start. I’ll give it a shot now this way.
Thank you.
Okay,
So I understand I cannot use Ff.
In response to Andrew~
3. Initial velocity impacts how fast you travel the distance, and then would impact the total time to travel, thus impacting force?
5. If you were given the time interval, would would work not just equal Δt ?
I believe there is a...
Homework Statement
A force varies with time according to the expression F=aΔt, where a = 2.0 N/s.
From this information, can you determine the work done on a particle that experienced this force over a displacement of 0.50 m?
Homework Equations
W = F*d
Vf = Vo + aΔt
F = ma
The Attempt at...
Friction changes signs when the force is applied in the opposite direction on the table block, and I would be lead to assume that that stays facing the opposite direction even after the force is no longer applied.
If that’s the case, then:
-a + g = 2a - .5g
3a = 1.5g
a = .5g
Which is correct...
Homework Statement
Assume that the block on the table (Figure 1) has twice the inertia of the hanging block.
(a)You give the block on the table a push to the right so that it starts to move. If the magnitude of the frictional force exerted by the table on the block is half the magnitude of...