Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Math says - double plate distance capacitor doubles capasity

Status
Not open for further replies.

Prototyp_V1.0

Advanced Member level 2
Joined
Apr 3, 2007
Messages
673
Helped
119
Reputation
238
Reaction score
83
Trophy points
1,308
Location
Norway
Activity points
5,103
capacitance formula distance

Hi.

Have this problem:

Have a capacitor with C = 5µF. What happens to the capacity when the distance between the plates doubles?

So the math. Formula for capacitance is:
\[C = 5 \mu F = \varepsilon_0 \cdot \varepsilon_r \frac{A}{d}\]

So the obvius thing (i think) is to multiple the d with 2, like this:
\[5 \mu F = {\varepsilon_0 \cdot \varepsilon_r} \frac{A}{2\cdot d}\]

Then I feel it's right to multiple both sides with 2, that gives:
\[10 \mu F = {\varepsilon_0 \cdot \varepsilon_r} \frac{A}{d}\]

But that is just obviously wrong, knowing that the more distance between the plates, the less will the capasity be. Simple question: What is wrong ?
 

capacitor math

Because d is the denominator, when you multiply d by 2, you are actually multiplying by one half.
Multiply both sides by one half and you get the intuitively obvious smaller capacitance.
 
capacitance coupling distance formulas

cherrytart said:
Because d is the denominator, when you multiply d by 2, you are actually multiplying by one half.
Multiply both sides by one half and you get the intuitively obvious smaller capacitance.
Ok, thanks. Guess that makes the chore correct :D
 

capacitor plates double distance

yea i agree
 

Re: Math says - double plate distance capacitor doubles capa

Prototyp_V1.0 said:
Hi.

Have this problem:

Have a capacitor with C = 5µF. What happens to the capacity when the distance between the plates doubles?

So the math. Formula for capacitance is:
\[C = 5 \mu F = \varepsilon_0 \cdot \varepsilon_r \frac{A}{d}\]

So the obvius thing (i think) is to multiple the d with 2, like this:
\[5 \mu F = {\varepsilon_0 \cdot \varepsilon_r} \frac{A}{2\cdot d}\]

Then I feel it's right to multiple both sides with 2, that gives:
\[10 \mu F = {\varepsilon_0 \cdot \varepsilon_r} \frac{A}{d}\]

But that is just obviously wrong, knowing that the more distance between the plates, the less will the capasity be. Simple question: What is wrong ?


heeeeey
when u multiply d by 2 u assume that the equ. reuslt is still 5 !! and u continue so one..

now u mult. d by 2 asif u divide the equ. by 2 so 5 is now 2.5

Ok

Thanks
 

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top