Capacitance & Potential Difference Question

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Homework Help Overview

The discussion revolves around a problem involving a parallel plate capacitor with a dielectric material. The capacitor has a capacitance of 5.2 µF and is charged by a 1.8-V battery. After disconnecting from the battery, the dielectric slab is removed, prompting questions about the resulting potential difference.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the charge on the capacitor and the effect of removing the dielectric slab on capacitance and potential difference. There are attempts to calculate charge and voltage using different equations, with some questioning the assumptions made in their calculations.

Discussion Status

The conversation includes various attempts to understand the relationship between charge, capacitance, and voltage. Some participants provide calculations and seek clarification on their reasoning, while others offer insights into the effects of the dielectric on capacitance. There is no explicit consensus on the final answer, but several productive lines of inquiry are being explored.

Contextual Notes

Participants are navigating the implications of the dielectric constant and its effect on capacitance when the dielectric is removed. There is also a mention of the charge remaining constant after disconnection from the battery, which is a key assumption in their calculations.

vbgirl
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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[tex]\Delta V[/tex]

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
 
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vbgirl said:
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?
 
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.
 
When the capacitor is first connected to the battery, what is the charge that ends up on the capacitor?
 
It should be 9.36e-6 coulomb
 
Last edited:
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?
 
It is decreased since the distance between the plates increases.
 
vbgirl said:
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)
 
Last edited:
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
 
  • #10
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?
 
  • #11
Not sure if this is right, but I get 9.36e-6/1.3e-6=7.2V
 
  • #12
If all the steps you took were right, how could the answer be anything else? :smile:

Your result is good.
 
  • #13
Thank you so much for the help, I really appreciate it.
 
  • #14
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:wink:
 
  • #15
oh lol .. new at this .. didnt even see the year when this was posted ! ;p
 

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