Math says - double plate distance capacitor doubles capasity

[Prototyp_V1.0]

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 ?

 

[cherrytart]

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.

 

[Prototyp_V1.0]

capacitance coupling distance formulas

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

 

[hsingarajah]

capacitor plates double distance

yea i agree

 

[ahmed osama]

Re: Math says - double plate distance capacitor doubles capa

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