Recent content by sw1mm3r

damn and i was tryin to figure out where i went wrong... kool sounds good i understand it then

1.2 is in the problem its 1.2 mm... and i have understood a good chunk of stuff i just got confused on the last part

no look c1/c2=d2/d1 (3.1x10^-11)/c2=.0045/.0012 c2=8.26x10^-12 so c2 is obviously not c1/3

thats the formula i was talking about

well by using your formula you do not get c2 being c1/3... you get c2=8.26x10^-12... and w=8.69 J But if its wrong can you give me the right answers as i have been grinding my gears on this problem??

ok i think i figured out on my own with a different easier method... where c2=8.26x10^-12 and W=8.69 J

Ok now i am super confused do you think someone could give me the value of the capacitance at 4.5 mm, because now i don't know how to solve for it

so then just to make sure i would have C=epsilon/d only

ok that makes sense so then in solving for the capacitance for 4.5 mm i would neglect A because its constant??

How can you use the C formula if you don't know area I found the electric field by using E=(K*Q)/d^2 C at 4.5 is 5.00x10^-13

what formula for C are you using? What U at 4.5 mm are you getting? yea that formula uses C to get a new C i think... our prof gave it to use in class... it gives the same values as C=Q/Ed

the problem gives you the capacitance at 1.5 so using U=(1/2)(Q^2)/C... I get U=3.16 and U at 4.5 mm gives me U=195.93 then taking the difference gives me Work=192.76 J and C=C*E/d is an equation our prof gave us, it gives the same values if you use C=Q/Ed

But i thought the question was asking for work not the capacitance? and I did not use that formula of e0*A/D because of the same reason we have no area which is why i used C=(C*E)/(d)... To find capacitance at 4.5 mm... Because they gave us capacitance at 1.5 mm in the problem...

idk maybe i missed something in lecture but can you explain why C after=C before*(1.2/4.5)

i thought you had to convert the mm to m