A parallel plate air-capacitor of capacitance C is subjected to potential difference V. What is the force between the plates of the capacitor if the separation of plates is d?
a)$$\frac{CV^2}{2d}$$
b)$$\frac{C^2V^2}{2d^2}$$
c)$$\frac{C^2V^2}{d^2}$$
d)$$\frac{C^2V}{d}$$

Question

A parallel plate air-capacitor of capacitance C is subjected to potential difference V. What is the force between the plates of the capacitor if the separation of plates is d?
a)$$\frac{CV^2}{2d}$$
b)$$\frac{C^2V^2}{2d^2}$$
c)$$\frac{C^2V^2}{d^2}$$
d)$$\frac{C^2V}{d}$$

anonymous · Answer

q = cv. f = q.q/ r^2

ujjwal · Answer

That doesn't help. And how do you get f=q.q/r^2?
I guess$$F=\frac{q\times q}{4\pi \epsilon_0 r^2}$$

ujjwal · Answer

@telltoamit ??

anonymous · Answer

oops sorry, then the answer shud be C2V2/4πϵ0d2

anonymous · Answer

anyways gtg, hope u find someone to help

ujjwal · Answer

ok.

anonymous · Answer

a) CV^2 /2d

anonymous · Answer

http://www.schoolphysics.co.uk/age16-19/Electricity%20and%20magnetism/Electrostatics/text/Force_between_two_charged_plates/index.html substitute Q = CV

ujjwal · Answer

How did you do it? But it says correct answer is d.

ujjwal · Answer

@Vaidehi09  Is that link reliable? well, i had also thought the same way earlier but then the answer is not correct probably..

anonymous · Answer

since the answer is not a....there must be something wrong....i'll try searching more.

anonymous · Answer

the force on the the plates is F = 1/2 QE........I just saw Prof. Lewin's lecture. so if we sub
E = V/d 
and Q = CV
then we should get F = CV^2/2d......this should be the answer.

ujjwal · Answer

yeah, may be my book is mis-printed...

ujjwal · Answer

what do you say @apoorvk ?

anonymous · Answer

here's the link for the lecture. listen to the first part. he explains why there is 1/2 in the formula. http://www.physicsforums.com/showthread.php?t=292158

apoorvk · Answer

All my latex work crashed D:

(btw option A is correct, let me just tell you where you're going wrong)

ujjwal · Answer

So we are going wrong and yet we are ending with correct answer? @apoorvk

anonymous · Answer

i'll check out ur explanation tomorrow morning. its quite late now, so gtg!

apoorvk · Answer

wait option A is NOT right am sorry.

ujjwal · Answer

Are you replying @apoorvk ? I am feeling sleepy...

apoorvk · Answer

Gimme a moment please, typing it all out again can get pretty boring.

ujjwal · Answer

Well walter lewin says Force between two parallel plates is:$$F=\frac{1}{2}qE$$so, i guess, it should be  (a).

apoorvk · Answer

Yeah. so...
C = Q/V = ϵoA/d
so, Q = CV = ϵoAV/d

Now, we calculate the electric field (E) between the capacitor plates.

Applying Gauss's Law to the positive plate, (let are of each face be 'A')
E'.(2A) = Q/ϵo = ϵoAV/dϵo = AV/d
or, E' = AV/2Ad = V/2d

Now for the contribution to the net electric field bet ween the two plates would be the same in magnitude as well as direction (i.e. towards the negative plate), so the net Elctric field between the plates = E = V/2d + V/2d = V/d

apoorvk · Answer

Now. what is the force acting upon??? The question asks about the "force acting between the  plates", but on which particle? The force would depend on the charge carried by the particle on which it acts!

apoorvk · Answer

Oh do you mean to ask about the attractive force between the two plates of the capacitor?

ujjwal · Answer

Its asking for the force between two plates as clearly stated in the question.

experimentx · Answer

not sure though
|dw:1341779549662:dw|