What Happens to the Voltage Across a Capacitor When the Plates are Separated?

[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
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?

 

[davenn]

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
Go to page 10
Voltage goes down . It is wrong . And formula is also wrong. Is that correct?

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 energyDave

 

[gracy]

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?

 

[nasu]

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

 

[SammyS]

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
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?

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 \$.