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Ch 25: The Electric Potential
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 25: The Electric PotentialProblem 22b
Chapter 25, Problem 22b

A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.0×105 V/m. How much charge is on each plate?

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Step 1: Recall the relationship between the electric field (E), the charge (Q), and the capacitance (C) of a parallel-plate capacitor. The electric field is given by \( E = \frac{Q}{\varepsilon_0 A} \), where \( \varepsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \ \text{F/m} \)), and \( A \) is the area of one plate.
Step 2: Calculate the area \( A \) of one plate using the formula for the area of a circle: \( A = \pi r^2 \), where \( r \) is the radius of the plate. The diameter is given as 3.0 cm, so the radius is \( r = \frac{3.0}{2} \ \text{cm} = 1.5 \ \text{cm} = 0.015 \ \text{m} \). Substitute this value into the formula for \( A \).
Step 3: Rearrange the formula for the electric field to solve for the charge \( Q \): \( Q = E \varepsilon_0 A \). Substitute the given electric field strength \( E = 1.0 \times 10^5 \ \text{V/m} \), the calculated area \( A \), and the permittivity of free space \( \varepsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \) into this equation.
Step 4: Perform the multiplication to find the charge \( Q \). Ensure that all units are consistent (meters, farads, volts, etc.) during the calculation.
Step 5: The result from the previous step gives the magnitude of the charge on each plate. Remember that the charge on one plate is positive, and the charge on the other plate is negative, as they are oppositely charged in a capacitor.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Capacitance

Capacitance is a measure of a capacitor's ability to store charge per unit voltage. It is defined as C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage across the plates. For parallel-plate capacitors, capacitance can also be calculated using the formula C = ε₀(A/d), where ε₀ is the permittivity of free space, A is the area of one plate, and d is the separation between the plates.
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Electric Field

The electric field (E) in a capacitor is the force per unit charge experienced by a positive test charge placed in the field. It is uniform between the plates of a parallel-plate capacitor and is given by E = V/d, where V is the voltage and d is the distance between the plates. The strength of the electric field is crucial for determining the potential difference and the charge on the plates.
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Charge on Plates

The charge (Q) on each plate of a capacitor can be calculated using the relationship Q = C × V. Once the capacitance is determined from the physical dimensions of the capacitor and the electric field strength, the voltage can be derived from the electric field and plate separation. This allows for the calculation of the total charge stored on each plate, which is equal in magnitude but opposite in sign.
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Textbook Question

Two 2.0-cm-diameter disks spaced 2.0 mm apart form a parallel-plate capacitor. The electric field between the disks is 5.0×105 V/m. An electron is launched from the negative plate. It strikes the positive plate at a speed of 2.0×107 m/s. What was the electron's speed as it left the negative plate?

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Textbook Question

Two 2.00 cm×2.00 cm plates that form a parallel-plate capacitor are charged to ±0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is 1.00 mm?

Textbook Question

A 3.0-cm-diameter parallel-plate capacitor has a 2.0 mm spacing. The electric field strength inside the capacitor is 1.0×105 V/m. What is the potential difference across the capacitor?