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