Capacitance & Potential Difference Question

[vbgirl]

Homework Statement

A parallel plate capacitor of capacitance 5.2 µF has the space between the plates filled with a slab of glass with κ = 4.0. The capacitor is charged by attaching it to a 1.8-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed.
Question: What is the potential difference after the glass is removed?

Homework Equations

Formula:
Q=Cx$$\Delta V$$

The Attempt at a Solution

Q= ( 5.2 X 1.8 )/2
I have gotten 4.68 but it is wrong. I am not sure what I am doing wrong

 

[gneill]

Q= ( 5.2 X 1.8 )/2
I have gotten 4.68 but it is wrong. I am not sure what I am doing wrong

Well, can you explain your strategy? What prompted the above equation?

 

[vbgirl]

Hmm should be Q= 1.8 C=5.2 and solve for V instead, for this question. But not sure how to solve for the difference.

 

[gneill]

When the capacitor is first connected to the battery, what is the charge that ends up on the capacitor?

 

[vbgirl]

It should be 9.36e-6 coulomb

 

[gneill]

Okay. Now, when the battery is disconnected, that charge remains on the capacitor.

When the dielectric slab is removed from between the plates, what happens to the capacitance value of the capacitor?

 

[vbgirl]

It is decreased since the distance between the plates increases.

 

[gneill]

It is decreased since the distance between the plates increases.

Not quite. The plates remain at the same separation, but the dielectric constant for air is different than that for the glass. Can you determine the value of the capacitance without the slab in place?

(HINT: it involves the dielectric constant, κ)

 

[vbgirl]

Would the capacitance become 1/4th of original since air is k= 1.0 to the initial k= 4.0 of glass? Resulting in C= 1.3e-6

 

[gneill]

That's right.

Now, the capacitor still has the same charge on it (it stays with the plates). So, given its (new) value and charge, what is the voltage across it?

 

[vbgirl]

Not sure if this is right, but I get 9.36e-6/1.3e-6=7.2V

 

[gneill]

If all the steps you took were right, how could the answer be anything else?

Your result is good.

 

[vbgirl]

Thank you so much for the help, I really appreciate it.

 

[soundslikeme]

okay so here's what i did ..
as q remains constant.. by the first data we get it as q=5.2 x 1.8 (as q=cv)
then .. as the dielectric slab is removed we get c as 5.8/4 [as k=c(medium)/c]
replace in the original equation we get v = 9.36/1.45 which is 6. 45 (as v=q/c and q remains contant)
so here u go .. acc. to me the answer is 6.45

 

[soundslikeme]

oh lol .. new at this .. didnt even see the year when this was posted ! ;p