The voltage across an air-filled parallel-plate capacitor is measured to be 173.0 V. When a dielectric is inserted and completely fills the space between the plates as in the figure below, the voltage drops to 35.3 V.

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The voltage across an air-filled parallel-plate capacitor is measured to be 173.0 V. When a dielectric is inserted and completely fills the space between the plates as in the figure below, the voltage drops to 35.3 V.

image

(a) What is the dielectric constant of the inserted material?


(b) Can you identify the dielectric?

    



(c) If the dielectric doesn't completely fill the space between the plates, what could you conclude about the voltage across the plates?

asked Jan 27, 2012 in Physics by apush ~Expert~ (552 points)
    

2 Answers

+2 votes
 
Best answer
For part A, you can find the dielectric constant by using the formula:

C = kC' (where C is the capacitance and k is the constant)

and then noting that C = Q/V (Q=charge, V=voltage)

We know that charge is constant, and that only the voltage changes, hence we can rewrite the first equation as:

V1*C = V2*k*C

Simplifying and solving for k we get:

k = V1 / V2 = 173 V / 35.3 V    =     4.9

k = 4.9

 

 

Part B,

a k value of 4.9 would be bakelite

(*other k values are as follows:    nylon = 3.4, paper = 3.7, neoprene rubber = 6.7, teflon = 2.1)

 

Part C,
 

Since the dielectric doesn't completely fill the space, that means the voltage is not at the minimal point it could be at.
answered Jan 29, 2012 by kirby ~Expert~ (3,020 points)
selected Aug 3, 2012 by pokemonmaster
0 votes
V won't reach the minimum value as when the dielectric is totally filling space within capacitor.

35.3 < Vpartially < 173
answered Apr 1, 2013 by anonymous

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