Question

The plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 80.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 4.00$\times$ 10${}^6$ V/m. What is (a) the potential difference between the plates; (b) the area of each plate; (c) the capacitance?

Short answer

(a) 10,000 V; (b) 2.26×10⁻² m²; (c) 8.01×10⁻¹² F.

Step-by-step solution

Finding Potential Difference

The potential difference (voltage) $V$ between the plates of the capacitor can be found using the formula for the electric field $E$ between two parallel plates: $$E=\frac{V}{d}$$where $d$ is the separation between the plates (2.50 mm or 0.00250 m). Rearranging the formula to solve for $V$, we have:$$V=E\times d$$Substitute $E=4.00\times {10}^6$ V/m and $d=0.00250$ m:$$V=4.00\times {10}^6\times 0.00250=10,000\text{ V}$$

Calculating Plate Area

The area $A$ of each plate can be found using the formula for the electric field between two plates:$$E=\frac{\sigma }{{\epsilon }_0}$$where $\sigma$ is the surface charge density and ${\epsilon }_0$ is the vacuum permittivity $8.85\times {10}^{-12}$ C²/(N·m²). First, find $\sigma$:$$\sigma =\frac{Q}{A}$$Set $\frac{Q}{A}=E\times {\epsilon }_0$:$$\frac{Q}{A}=4.00\times {10}^6\times 8.85\times {10}^{-12}$$$$A=\frac{Q}{4.00\times {10}^6\times 8.85\times {10}^{-12}}$$Substituting $Q=80.0\times {10}^{-9}$ C:$$A=\frac{80.0\times {10}^{-9}}{4.00\times {10}^6\times 8.85\times {10}^{-12}}=2.26\times {10}^{-2}{\text{ m}}^2$$

Determining Capacitance

The capacitance $C$ of a parallel-plate capacitor is given by:$$C=\frac{{\epsilon }_0\cdot A}{d}$$Substitute in the known values of ${\epsilon }_0=8.85\times {10}^{-12}$ C²/(N·m²), $A=2.26\times {10}^{-2}{\text{ m}}^2$, and $d=0.00250\text{ m}$:$$C=\frac{8.85\times {10}^{-12}\times 2.26\times {10}^{-2}}{0.00250}=8.01\times {10}^{-12}\text{ F}$$

Conclusion

We've found the potential difference is 10,000 V, the plate area is $2.26\times {10}^{-2}$ m², and the capacitance is $8.01\times {10}^{-12}$ F.

Key concepts

Electric Field

An electric field is a region around a charged object where other charged objects experience a force. In a parallel-plate capacitor, the electric field is uniform between the two plates.

This means the electric field strength is constant across the entire space between the plates. It is this field that enables the capacitor to store potential energy, as it exerts a force on charges within its vicinity.

For a parallel-plate capacitor, the strength of the electric field (E) is given by the equation:$$E=\frac{V}{d}$$where V is the potential difference between the plates, and d is the distance between them.

The electric field is measured in volts per meter (V/m) and its value depends on both the potential difference and the separation distance between the plates. In our specific example, the electric field is known to be 4.00 \times 10^6 V/m, giving us a basis to solve for other properties of the capacitor.

Potential Difference

The potential difference, also known as voltage, is crucial in the context of a capacitor. It represents the energy per unit charge that is stored across the capacitor plates. This difference arises because of the work done to move a charge between the plates.

In simpler terms, the greater the potential difference, the more energy the capacitor can store. You can find the potential difference using the formula:$$V=E\times d$$where E is the electric field, and d is the distance between the plates.

The potential difference is measured in volts (V). In our solved problem, we computed the potential difference to be 10,000 V using the given electric field and plate separation. This high voltage reflects how efficiently the capacitor can store energy when the electric field is strong and the plates are sufficiently close.

Capacitance

Capacitance is the measure of a capacitor’s ability to store charge. It is defined as the amount of charge a capacitor can store per unit potential difference. This property is crucial because it indicates how much electrical energy the capacitor can hold.

It is given by the formula:$$C=\frac{{\epsilon }_0\cdot A}{d}$$where \varepsilon_0 is the vacuum permittivity, A is the area of one of the plates, and d is the distance between the plates.

Capacitance is measured in farads (F). In our example, after substituting the known values, the calculated capacitance is 8.01 \times 10^{-12} F, or slightly over 8 picofarads (pF). This value tells us that the capacitor can store a small amount of charge, appropriate for the given area and separation distance in a vacuum.

Surface Charge Density

Surface charge density (\sigma) represents how much charge is accumulated on the surface of either plate of a capacitor. It is an essential parameter because it affects the electric field's magnitude and, consequently, the energy stored in the capacitor.

The relationship between the electric field and surface charge density is expressed as:$$E=\frac{\sigma }{{\epsilon }_0}$$where \varepsilon_0 is the vacuum permittivity. From this, we can derive \sigma as a function of known variables, such as the electric field and vacuum permittivity.

In our specific problem, by using the known electric field strength and permittivity, we determine the surface charge density. This lets us infer how effectively the capacitor plates can handle the electric field generated by the stored charge. Understanding this concept helps appreciate the plate's capability in holding certain charge levels, crucial for practical applications.