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1. ### Potential across a conducting sphere surrounded by an insulator

Re- ran through the calculations got .47 (.465). Thank you so much for your help. You guys are always great. Edit: yeah, just a typo
2. ### Potential across a conducting sphere surrounded by an insulator

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...
3. ### Potential across a conducting sphere surrounded by an insulator

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 ...
4. ### Potential across a conducting sphere surrounded by an insulator

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...
5. ### E-field from one end of an infinite non-conducting rod

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.
6. ### E-field from one end of an infinite non-conducting rod

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.
7. ### E-field from one end of an infinite non-conducting rod

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...
8. ### E-field from one end of an infinite non-conducting rod

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?
9. ### E-field from one end of an infinite non-conducting rod

Okay, I tried solving for qo via: dq = -lqoe-x/l 2.9μC = ∫0∞-lqoe-x/l 2.9*10-6 = lqo qo = 2.9*10-6/(.0286)
10. ### E-field from one end of an infinite non-conducting rod

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...
11. ### E-field from one end of an infinite non-conducting rod

Homework Statement Homework Equations dE= k dq/r2 The Attempt at a Solution [/B] 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...
12. ### Can we find work done on this particle, and if not, what are we missing?

Cool. Thanks again! I feel like I have a much better understanding of work and using calculus in physics! -arturo
13. ### Can we find work done on this particle, and if not, what are we missing?

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...
14. ### Can we find work done on this particle, and if not, what are we missing?

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.
15. ### Can we find work done on this particle, and if not, what are we missing?

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...
16. ### Can we find work done on this particle, and if not, what are we missing?

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...
17. ### Half Atwood machine: acceleration after inital force

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...
18. ### Half Atwood machine: acceleration after inital force

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...