Question

Which of the following capacitors has the largest charge? a) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery b) a parallel plate capacitor with an area of \(5 \mathrm{~cm}^{2}\) and a plate separation of \(1 \mathrm{~mm}\) connected to a \(10-\mathrm{V}\) battery c) a parallel plate capacitor with an area of \(10 \mathrm{~cm}^{2}\) and a plate separation of \(4 \mathrm{~mm}\) connected to a \(5-\mathrm{V}\) battery d) a parallel plate capacitor with an area of \(20 \mathrm{~cm}^{2}\) and a plate separation of \(2 \mathrm{~mm}\) connected to a \(20-\mathrm{V}\) battery e) All of the capacitors have the same charge.

Short answer

A) Area: 1 cm^2, Plate separation: 2 mm, Battery: 10 V B) Area: 0.5 cm^2, Plate separation: 1 mm, Battery: 10 V C) Area: 1 cm^2, Plate separation: 4 mm, Battery: 5 V D) Area: 20 cm^2, Plate separation: 2 mm, Battery: 20 V Answer: D) Area: 20 cm^2, Plate separation: 2 mm, Battery: 20 V

Step-by-step solution

Recall the formula for the charge of a parallel plate capacitor

The formula for the charge of a parallel plate capacitor is given by \(Q=\epsilon _0 A \frac{V}{d}.\)

Convert units

Before we calculate the charges, we need to convert the given areas from \(\mathrm{cm}^2\) to \(\mathrm{m}^2,\) and the plate separations from \(\mathrm{mm}\) to \(\mathrm{m}.\) The conversion factors are: 1 cm = 0.01 m; 1 mm = 0.001 m.

Calculate the charge for each capacitor

The vacuum permittivity is given by \(\epsilon _0 = 8.854\times10^{-12}\mathrm{~F~m}^{-1}\). Calculate the charges for each capacitor using the formula and the converted units: a) \(Q_a = \epsilon_0 A_a \frac{V_a}{d_a} = (8.854\times10^{-12})(0.001) \frac{10}{0.002} = 44.27\times10^{-12}\mathrm{C}\) b) \(Q_b = \epsilon_0 A_b \frac{V_b}{d_b} = (8.854\times10^{-12})(0.0005) \frac{10}{0.001} = 44.27\times10^{-12}\mathrm{C}\) c) \(Q_c = \epsilon_0 A_c \frac{V_c}{d_c} = (8.854\times10^{-12})(0.001) \frac{5}{0.004} = 11.0675\times10^{-12}\mathrm{C}\) d) \(Q_d = \epsilon_0 A_d \frac{V_d}{d_d} = (8.854\times10^{-12})(0.002) \frac{20}{0.002} = 177.08\times10^{-12}\mathrm{C}\)

Compare the charges

Compare the charges: \(Q_a = 44.27\times10^{-12}\mathrm{C}\) \(Q_b = 44.27\times10^{-12}\mathrm{C}\) \(Q_c = 11.0675\times10^{-12}\mathrm{C}\) \(Q_d = 177.08\times10^{-12}\mathrm{C}\) The largest charge is found in option d) with a value of \(177.08\times10^{-12}\mathrm{C}\). Therefore, the parallel plate capacitor with an area of \(20\mathrm{~cm}^2\) and a plate separation of \(2\mathrm{~mm}\) connected to a \(20-\mathrm{V}\) battery has the largest charge.

Key concepts

Vacuum Permittivity

Vacuum permittivity, also known as the electric constant or permittivity of free space, is a fundamental physical constant denoted by the symbol \( \epsilon_0 \). This constant is vital in electromagnetic theory, particularly in the context of capacitors. It describes how much electric field is 'permitted' to exist in a vacuum before it affects surrounding charges.
For parallel plate capacitors, \( \epsilon_0 \) plays a crucial role in calculating the capacitance and, subsequently, the charge stored. The value of vacuum permittivity is approximately \( 8.854 \times 10^{-12} \mathrm{~F~m}^{-1} \) (farads per meter).
This constant is essential in the equation \( Q = \epsilon_0 A \frac{V}{d} \), allowing us to determine the charge \( Q \) based on the area \( A \), potential difference \( V \), and plate separation \( d \). Understanding \( \epsilon_0 \) is fundamental for anyone studying electromagnetism and related fields.

Charge Calculation

To find the charge stored in a parallel plate capacitor, we use the formula:

• \( Q = \epsilon_0 A \frac{V}{d} \)

In this formula, \( Q \) represents the charge, \( \epsilon_0 \) is the vacuum permittivity, \( A \) is the area of one of the plates, \( V \) denotes the voltage (or potential difference) across the plates, and \( d \) is the distance between the plates.
The larger the area and voltage, the greater the charge the capacitor can store. Conversely, the larger the distance between the plates, the less charge it can store, as this decreases the electric field generated across the plates.
By substituting the values for each parameter into this expression, students can accurately compute the charge stored in different configurations of capacitors, helping them grasp how each variable influences the result.

Capacitance Formula

The capacitance formula is at the heart of understanding how capacitors work, especially in parallel plate configurations. Capacitance \( C \) is defined as the ratio of the charge \( Q \) on each plate to the voltage \( V \) across the plates:

• \( C = \epsilon_0 \frac{A}{d} \)

This equation highlights how capacitance depends not only on material constants but also on geometric factors — specifically the plate area \( A \) and the separation distance \( d \).
With a greater area, capacitance increases because the capacitor can hold more charges. A decrease in the distance \( d \) also boosts capacitance, as it enables a stronger electric field between the plates, allowing for more charge storage through a stronger attraction. This formula provides a groundwork for calculating capacitance and, when tied with the equation for charge (\( Q = CV \)), can lead to deeper investigations into electric fields.

Unit Conversion

Unit conversion is a necessary step in solving physics problems accurately. With capacitors, area and distance are often given in different units, which need to be consistent when applying formulas.
For example, if the area is given in \( \mathrm{cm}^2 \), convert it to \( \mathrm{m}^2 \) by multiplying by \(0.0001\), as 1 cm = 0.01 m and hence 1 \( \mathrm{cm}^2 = 0.0001 \mathrm{m}^2 \). Similarly, convert millimeters to meters for plate separation, using 1 mm = 0.001 m.

• Area conversion: \( ext{cm}^2 \rightarrow ext{m}^2 \)
• Distance conversion: mm \rightarrow m

Without proper unit conversion, the calculations would yield incorrect results due to inconsistent units. Understanding unit conversion is crucial not only in capacitor problems but in all aspects of scientific calculation.