# Capacitance pdf crosscheck

1. Feb 23, 2016

### gracy

A capacitor has a capacitance C and is charged using a battery. After disconnecting the battery the two plates are separated until they are twice as far apart as originally. What happens to the voltage across the capacitor?
I think answer should be voltage increases . As battery has been disconnected q will remain same . Distance between the plates is increasing , capacitance should decrease.
$Q$= CΔV
As Q is constant and C is decreasing , V should increase.
But according to http://www.bama.ua.edu/~tmewes/PH-106/2008/clicker%20questions/CH26%20Clicker.pdf [Broken]
Go to page 10
Voltage goes down . It is wrong . And formula is also wrong. Is that correct?

I have gone through all 14 pages of this pdf and found them right except this one , but I want to confirm are all other 13 pages correct?

Last edited by a moderator: May 7, 2017
2. Feb 23, 2016

### davenn

I agree with you ... it may have been a genuine highlighting mistake ?

it increases because you do work on the system to spread the plates, this results in an increase in potential energy

Dave

Last edited by a moderator: May 7, 2017
3. Feb 23, 2016

### gracy

4. Feb 23, 2016

### nasu

Yes, the formula is wrong on that page.(10)

5. Feb 24, 2016

### SammyS

Staff Emeritus
You are right.

Solving $\ Q= C\,ΔV\$ for ΔV gives $\displaystyle\ \Delta V = \frac QC \$.

That pdf / PowerPoint has the incorrect $\displaystyle\ \Delta V = \frac CQ \$.

Last edited by a moderator: May 7, 2017