BR Physics, Section 9, REVIEW QUESTIONS, Question 22

[sillyjoe]

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The question reads:

What happens over time to the capacitance of a capacitor, as the space between its plates is filled at a constant rate with a polar solution?

A. The capacitance remains constant until the volume is completely filled, then the capacitance immediately drops to the new value of K*C0.
B. The capacitance remains constant until the volume is completely filled, then the capacitance immediately increases to the new value of K*C0.
C. The capacitance gradually decreases as the solution is added until the volume is completely filled, then the capacitance remains at the new value of K*Co
D. The capacitance gradually increases as the solution is added until the volume is completely filled, then the capacitance remains at the new value of K*C0

Choice D is the best answer. As the space between the capacitor plates is filled with a polar substance, the capacitance will increase. This eliminates choices A and C. Because the fluid is being added at a constant rate, the capacitance will change at a constant (gradual) rate, not abruptly. The best answer is choice D.

My confusion I think stems from a lack of conceptual understanding as to how a capacitor works and how placing objects between the capacitor effects it. First off, does anyone have a good video or video series to explain this and other electricity concepts? I tried Khan, but I just found it too dumbed down...

Anyways, I was under the assumption that you can increase capacitance by placing a dielectric in between the plates. However, when the dielectric starts to ionize you will get dielectric breakdown. I then assumed that a polar material would lead to a breakdown?

Clearly I am not understanding this so any help would be appreciated!

 

[gettheleadout]

A polarizable dielectric only establishes surface charge density when placed within an electric field, unless the electric field is sufficiently strong for dielectric breakdown to occur where the dielectric acts as a conductor. You are correct that capacitance is increased with the inclusion of a dielectric in the field.

What's important to know is that a capacitor stores energy in the form of electric potential energy by establishing separation of charge. Capacitance represents the degree to which is possible such that C = Q / ∆V, but capacitance does not depend on the charge or voltage difference per se. The factors that actually determine capacitance are the physical features of the capacitor itself; for parallel plate capacitors, this means the area of the plates and their separation (d) such that C = εA / d.

For high-symmetry systems such as parallel plate capacitors (PPC's), the inclusion of a dielectric or conductor inside the field only alters the field within the dielectric or conductor. You should know that no electric field can exist within a conductor, and it's clear that if you completely filled the capacitor space with a conductor you would have the equivalent of a wire, so that isn't helpful for increasing capacitance. If you add a slice of conductor to one plate of a PPC and gradually make it thicker, you'll increase capacitance but only because you're effectively decreasing d, the plate separation. Unfortunately capacitors have something called a breakdown voltage, meaning you can only get the plates so close and the capacitance so high this way before the electric potential difference is so great that the plates will spark and discharge (short) the capacitor. So how else can we increase the capacitance?

Adding an insulating material (a dielectric) between the plates will result in a decreased electric field (the density of the field lines inside the dielectric is reduced by a factor of K, the dielectric constant where K = E / E_diel) within the dielectric. This is because, as an insulator, the dielectric material responds to the application of an external electric field with induced charge separation inside the material. For a dielectric in a uniform electric field (such as between the plates of a PPC) this means a surface charge density of opposing sign on either side-surface of the dielectric. The more polarizable the dielectric, the more charge separation can occur within it and the weaker the field inside will be as the field lines are thinned more inside relative to the external field.

Now, this becomes somewhat difficult to describe, but as you fill up more of the interplate space with dielectric the volume of the space that is left with the original electric field decreases. This results in a lower average electric field throughout the entire volume of the interplate space, and because the electric potential difference ∆V is directly proportional to the field strength E and the separation d (∆V = Ed for a uniform electric field), the value of the overall ∆V for the PPC decreases. Since C = Q / ∆V this increases the capacitance until the the dielectric completely fills the interplate space, and instead of having to solve for ∆V the complicated way by dividing the regions of the interplate space and solving for field components and stuff, the calculation simplifies to yield C = KC_0 = KεA / d.