iTeachChem•2mo ago

Capacitor 4

F = 1/2(dC/dx)E²
This will be the force that causes the springs to compress. Then how do i relate all of this ?

iTeachChem Helper•10/25/25, 12:25 PM

@Gyro Gearloose

iTeachChem Helper•10/25/25, 12:25 PM

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AY•10/25/25, 12:35 PM

for part a.) use energy conservation

AY•10/25/25, 12:36 PM

before the voltage was appliead, capacitor had zero potential energy

AY•10/25/25, 12:36 PM

also note that 3 springs are in parallel so equivalent spring constant is 3k

AY•10/25/25, 12:36 PM

-1/2 * 3k * (d0)^2 = E(initial)

AY•10/25/25, 12:37 PM

after this, we apply a potential and distance changes

AY•10/25/25, 12:37 PM

AY•10/25/25, 12:41 PM

1/2 * CV^2 - 1/2 * (3k) * (d)^2

AY•10/25/25, 12:41 PM

final energy

AY•10/25/25, 12:41 PM

equate them

AY•10/25/25, 12:41 PM

and write C = A(E0)/d

AY•10/25/25, 12:41 PM

u will get V

AY•10/25/25, 12:41 PM

also the approximation

AY•10/25/25, 12:42 PM

d0 << sqrt(A)

AY•10/25/25, 12:42 PM

d0^2 << A

AY•10/25/25, 12:42 PM

d0^2/A <<< 1

AY•10/25/25, 12:42 PM

simply neglect it

AY•10/25/25, 12:43 PM

u should get (3kd^3/AE0)^1/2 = V

PrasanOP•10/25/25, 12:54 PM

the answer isnt this

PPrasan the answer isnt this

AY•10/25/25, 12:55 PM

ok before i check

AY•10/25/25, 12:55 PM

for part b use energy method

PrasanOP•10/25/25, 12:55 PM

i havent seen part b yet

AY•10/25/25, 12:55 PM

afaik this is to be done by energy conservation

AY•10/25/25, 12:55 PM

not force

PrasanOP•10/25/25, 12:56 PM

u sayin for part b use energy method ?

PrasanOP•10/25/25, 12:56 PM

for time period?

AY•10/25/25, 12:56 PM

yes

AAY 1/2 * CV^2 - 1/2 * (3k) * (d)^2

AY•10/25/25, 12:57 PM

diff this = const

PrasanOP•10/25/25, 12:57 PM

i think if we get the expression of force and compare it with a = w^2x, we can get w

PrasanOP•10/25/25, 12:57 PM

agar aisa hua to

PrasanOP•10/25/25, 12:58 PM

apna bataio ki energy cons se kese time period?

PPrasan the answer isnt this

AY•10/25/25, 12:59 PM

what is the ans u got

PrasanOP•10/25/25, 12:59 PM

whole root 6kd^2(d0-d)/AE0

AY•10/25/25, 1:00 PM

gimme a min

PrasanOP•10/25/25, 1:00 PM

i think youve done it right

AY•10/25/25, 1:04 PM

without appxmation im getting sqrt(3k(d0^2 - d^2)d/AE0)

PrasanOP•10/25/25, 1:05 PM

yeah im getting the same also

AY•10/25/25, 1:05 PM

d0 - d...

PrasanOP•10/25/25, 1:05 PM

now the approximation that they have given us

PrasanOP•10/25/25, 1:05 PM

is with area

AY•10/25/25, 1:05 PM

yes

PrasanOP•10/25/25, 1:05 PM

the only way to use that is how u have done it

PrasanOP•10/25/25, 1:05 PM

i think

AY•10/25/25, 1:05 PM

wait i think i have something

AY•10/25/25, 1:09 PM

welp no

Capacitor 4

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