Chapter 4: Q61P (page 813)

A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates. (a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material? (b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?

Short Answer

Expert verified

(a) $K = 3.91$ .

(b) $V_{n} = 22.84 V$ .

Step by step solution

Parameters.

$\begin{array}{l} V_{0} = 45 V \\ V = 11.5 V \end{array}$

(a) Computing the dielectric constant of the material.

The constant of the dielectric material is given by

$K = \frac{V_{0}}{V}$

Pug the values in the equation

$K = \frac{V_{0}}{V} = \frac{45}{11.5} = 3.91 .$

(b) Finding the voltmeter read.

Now only the third of the area is filled with dielectric material and the rest is air so the capacitance is considered as two separate capacitor in parallel and the new capacitance is given by

$\begin{array}{l} C_{n} = C_{1} + C_{2} \\ = \frac{\in A}{3 d} + \frac{2 \in_{0} A}{3 d} \\ = \frac{K \in_{0} A}{3 d} + \frac{2 \in_{0} A}{3 d} \\ = \frac{\in_{0} A}{d} (\frac{3.91}{3} + \frac{2}{3}) \\ = \frac{1.97 \in_{0} A}{d} = 1.97 C_{0} \end{array}$

The charge is constant so

$C_{0} V_{0} = C_{n} V_{n}$ (1)

Substituting in (1) yields

$C_{0} (45) = 1.97 C_{0} V_{n} V_{n} = 22.84 V .$

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Short Answer

Step by step solution

Parameters.

(a) Computing the dielectric constant of the material.

(b) Finding the voltmeter read.

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