Understanding the Effects of Dielectrics on Capacitance: A Scientific Approach

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Homework Help Overview

The discussion revolves around the effects of inserting a dielectric into a charged capacitor after disconnecting it from a battery. Participants explore the implications for voltage, potential energy, and capacitance, which are central concepts in electrostatics.

Discussion Character

  • Conceptual clarification, Assumption checking, Exploratory

Approaches and Questions Raised

  • Participants discuss the validity of statements regarding voltage, potential energy, and capacitance when a dielectric is introduced. Questions arise about the behavior of charge and voltage in the absence of the battery, as well as the relationship between electric field and capacitance.

Discussion Status

Some participants have provided insights into the behavior of voltage and charge when a dielectric is inserted, while others have questioned the assumptions underlying the problem. There is an ongoing exploration of how the dielectric constant affects capacitance and electric field, with no explicit consensus reached.

Contextual Notes

Participants are navigating the implications of a disconnected battery on the capacitor's charge and voltage, as well as the definitions and properties of dielectrics. The discussion reflects a mix of understanding and uncertainty regarding these concepts.

RoboNerd
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Homework Statement



A capacitor, is fully charged by a battery. The battery is disconnected and a dielectric is inserted into the capacitor. Which of the following is true?

I. Voltage will stay the same
II. Potential energy of the capacitor will increase
III. capacitance will increase

Homework Equations


U = 0.5 * Q * deltaV

The Attempt at a Solution



I know that III is true as dielectrics by definition increase capacitance. How can I eliminate the other two to get the final answer of III only?

Thanks in advance.
 
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Battery maintains constant voltage.Since it is removed voltage will change.
But there is no flow of charge.Using the relation between Charge ,Potential and capacitance and this information find change in voltage and potential energy.
 
Technically you don't need to "eliminate" the other choices as long as you are certain that a specific choice is correct in a multiple choice question.

But I) is wrong simply because inserting a dielectric changes the electric field E and it is V=Ed where d the (constant) distance between the plates.

II) you can answer it yourself, using the equation you wrote in OP and calculating how voltage will change (increase or decrease)? (Charge will remain the same (why?))
 
Delta² said:
II) you can answer it yourself, using the equation you wrote in OP and calculating how voltage will change (increase or decrease)? (Charge will remain the same (why?))

Charge will remain the same because the battery is not connected to the capacitor, so the battery can not pump out more charged particles in response to the insertion of the dielectric.

harsh_sinha said:
Battery maintains constant voltage.Since it is removed voltage will change.
But there is no flow of charge

Why does a battery maintain constant voltage? And, I take it that there is no flow of charge because there is no potential difference caused by having a battery hooked up to the circuit?
 
RoboNerd said:
Charge will remain the same because the battery is not connected to the capacitor, so the battery can not pump out more charged particles in response to the insertion of the dielectric.
Why does a battery maintain constant voltage? And, I take it that there is no flow of charge because there is no potential difference caused by having a battery hooked up to the circuit?
Battery is like a pumping device. It pumps charge.When connected to a capacitor it pumps charge till steady state is achieved.In steady state when there is no flow of charge we can easily prove that potential difference across the capacitor is constant using Kirchoff's Law.
 
I wonder what have you been taught about the dielectric constant k:

1) A capacitor with dielectric of dielectric constant k, has capacitance ##C_k=kC_0## where ##C_0## the Capacitance with vacuum between its plates.
or
2) The electric field between the plates of a capacitor with dielectric k is ##E_k=\frac{E_0}{k}## where ##E_0## the electric field with vacuum between the plates.
 
Delta² said:
I wonder what have you been taught about the dielectric constant k:

1) A capacitor with dielectric of dielectric constant k, has capacitance Ck=kC0Ck=kC0C_k=kC_0 where C0C0C_0 the Capacitance with vacuum between its plates.
or
2) The electric field between the plates of a capacitor with dielectric k is Ek=E0kEk=E0kE_k=\frac{E_0}{k} where E0E0E_0 the electric field with vacuum between the plates.

yes, I know how a dielectric works. thank you.

I see how I should solve this problem. Thanks a lot everyone!
 

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