Question

Inspector Clouseau has found a clue in his pursuit of The Phantom:5 a sheet of some
unknown material. He has a parallel plate capacitor and can measure the maximum
charge held by the plates. With air between them, this maximum charge is 23.9 pC.
He then slips the unknown material between them and finds a maximum charge of
134 pC. Using this information, find the dielectric constant of the unknown material
and determine its identity using the Table 16.1 in your text.

Answer 1

Answer on Question 65309, Physics, Electric Circuits

Question:

Inspector Clouseau has found a clue in his pursuit of The Phantom 5: a sheet of some unknown material. He has a parallel plate capacitor and can measure the maximum charge held by the plates. With air between them, this maximum charge is $23.9\,pF$. He then slips the unknown material between them and finds a maximum charge of $134\,pF$. Using this information, find the dielectric constant of the unknown material and determine its identity using the Table 16.1 in your text.

Solution:

Let's first write the charge stored on the plates of both capacitors:

$$Q_{1max} = C_1 V,$$ $$Q_{2max} = C_2 V,$$

here, $Q_{1max}$ is the maximum charge stored on the plates of air-filled capacitor; $Q_{2max}$ is the maximum charge stored on the plates of the capacitor filled with unknown material; $C_1$ is the capacity of air-filled capacitor; $C_2$ is the capacity of the capacitor filled with unknown material and $V$ is the voltage across the plates of the capacitor (we assume that we connect the same voltage source to both capacitors).

Then, we can rewrite the expressions for $Q_{1max}$ and $Q_{2max}$:

$$Q_{1max} = C_1 V = \varepsilon_0 \frac{A}{d} V,$$ $$Q_{2max} = C_2 V = \kappa \varepsilon_0 \frac{A}{d} V,$$

here, $\kappa$ is the dielectric constant of the unknown material, $\varepsilon_0$ is the permittivity of free space, $A$ is the area of plate overlap, $d$ is the plate separation.

Let's express $V$ from both expressions:

$$V = \frac{Q_{1max}}{\varepsilon_0 A} d,$$ $$V = \frac{Q_{2max}}{\kappa \varepsilon_0 A} d.$$

Since $V$ is the same we can equate both expressions:

$$\frac {Q _ {1 \max}}{\varepsilon_ {0} A} d = \frac {Q _ {2 \max}}{\kappa \varepsilon_ {0} A} d,$$ $$Q _ {1 \max} = \frac {Q _ {2 \max}}{\kappa}.$$

From the last expression we can find the dielectric constant of the unknown material:

$$\kappa = \frac {Q _ {2 \max}}{Q _ {1 \max}} = \frac {134 \, pF}{23.9 \, pF} = 5.6.$$

Unfortunately, there is no table attached to this question, but knowing the dielectric constant you can determine the identity of the unknown material using your textbook.

Answer:

$$\kappa = 5.6.$$

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