Question

A parallel-plate capacitor is connected to a battery that maintains a constant potential difference between the plates. If the spacing between the plates is doubled, how is the magnitude of charge on the plates affected?

Answer 1

for parallel plate capacitors capacitence is
\[C=\frac{ e_0*A }{ d }\]
where
e_0=permittivity of the medium between the plates
A=area
d=distence between the plates

Answer 2

again we know that simply
\[C=Q/V\]
ok??

Answer 3

they are asking to give the answer as Qnew/Qold

Answer 4

that is what is confusing me. usually they ask for an answer "in terms of" but i just have a problem figuring out how to answer the question

Answer 5

so we've got two equations
\[C=\frac{ e_0*A }{ d}........................(1)\]and
\[C=\frac{ Q }{ V }.......................................(2)\]now
equate them
\[\frac{ E_0*A }{ d }=\frac{ Q }{V}\]
so from this equation we can see that Q=charge is inversely proportional to the distence between the plates\[Q \alpha (1/d)\]
\[\frac{ Q_(new) }{ Q_(old) }=\frac{ d_{old} }{ d_{new} }\]

Answer 6

got this?

Answer 7

let me test it out!

Answer 8

it said that was not correct

Answer 9

Incorrect; Try Again; 3 attempts remaining
The correct answer does not depend on: dnew, dold.

Answer 10

no it will surely not depend upon dnew and dold u have to substitute values of them
\[\frac{ Q_{new} }{ Q_{old}}=\frac{ d_{old} }{ 2*d_{old}}=1/2\]

Answer 11

so
\[Q_{new}=\frac{ Q_{old} }{ 2}\]

Answer 12

now try this

Answer 13

i entered .5 and it was correct

Answer 14

tricky!!! I hate it when I cant figure out what terms to answer a question in!

Answer 15

lol :)