A student conducted a study on parallel-plate capacitors to determine how capacitance is a… — ACT

A student conducted a study on parallel-plate capacitors to determine how capacitance is affected by physical dimensions. In Experiment 1, the student set the distance between the plates to a constant value $d_1$ and measured the capacitance ( $C$ , in picofarads, $\text{pF}$ ) for plates of various surface areas ( $A$ , in $\text{cm}^2$ ). The results are shown in Table 1.

Plate Area $A$ ( $\text{cm}^2$ )	Capacitance $C$ ( $\text{pF}$ )
10.0	8.8
20.0	17.6
30.0	26.4
40.0	35.2

In Experiment 2, the student used plates of a constant surface area $A_1$ and measured the capacitance at various plate separation distances ( $d$ , in millimeters, $\text{mm}$ ). The results are shown in Table 2.

Plate Separation $d$ ( $\text{mm}$ )	Capacitance $C$ ( $\text{pF}$ )
1.0	35.2
2.0	17.6
4.0	8.8
8.0	4.4

Based on these results, if the student constructs a capacitor using the same materials with a plate area of $60.0\text{ cm}^2$ and a plate separation distance of $4.0\text{ mm}$ , what will be the expected capacitance of this capacitor?

A
$5.9\text{ pF}$
$13.2\text{ pF}$ Answer
C
$93.9\text{ pF}$
D
$211.2\text{ pF}$

Answer

The expected capacitance of the capacitor is

13.2\text{ pF}

The data shows that capacitance is directly proportional to plate area and inversely proportional to plate separation distance. Starting from a baseline of

A = 40.0\text{ cm}^2

and

d = 1.0\text{ mm}

(where

C = 35.2\text{ pF}

), increasing the area to

60.0\text{ cm}^2

(a factor of

1.5

) increases the capacitance to

52.8\text{ pF}

. Then, increasing the separation distance to

4.0\text{ mm}

(a factor of

4

) divides the capacitance by

4

, resulting in

13.2\text{ pF}

Step-by-Step Solution

Determine the proportional relationship between capacitance and plate area from Experiment 1.

Capacitance (

C

) is directly proportional to plate area (

A

) because doubling the area (e.g., from

10.0\text{ cm}^2

20.0\text{ cm}^2

) doubles the capacitance (from

8.8\text{ pF}

17.6\text{ pF}

), meaning

C \propto A

Establishing the relationship for the first independent variable is necessary to scale its effect.

Determine the proportional relationship between capacitance and plate separation distance from Experiment 2.

Capacitance (

C

) is inversely proportional to plate separation (

d

) because doubling the distance (e.g., from

1.0\text{ mm}

2.0\text{ mm}

) halves the capacitance (from

35.2\text{ pF}

17.6\text{ pF}

), meaning

C \propto \frac{1}{d}

Establishing the relationship for the second independent variable is necessary to scale its effect.

Select a baseline configuration from the data to perform the scaling calculation.

Using the configuration from Table 2 where

d = 1.0\text{ mm}

and

C = 35.2\text{ pF}

. From Table 1, we see this corresponds to a constant plate area

A_1 = 40.0\text{ cm}^2

A known reference point with both dimensions and capacitance is needed as a starting point.

Scale the baseline capacitance for the change in plate area from

40.0\text{ cm}^2

60.0\text{ cm}^2

The area increases by a factor of

\frac{60.0}{40.0} = 1.5

. Since

C

is directly proportional to

A

, the capacitance increases to

35.2\text{ pF} \times 1.5 = 52.8\text{ pF}

d = 1.0\text{ mm}

This accounts for the direct scaling of the plate area.

Scale the intermediate capacitance for the change in plate separation from

1.0\text{ mm}

4.0\text{ mm}

The separation distance increases by a factor of

\frac{4.0}{1.0} = 4

. Since

C

is inversely proportional to

d

, the capacitance is divided by

4

, yielding

\frac{52.8\text{ pF}}{4} = 13.2\text{ pF}

This accounts for the inverse scaling of the plate separation distance.

Key Concept

Direct and Inverse Proportionality in Experimental Data

Question

Answer

Step-by-Step Solution

Key Concept

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