Skip to main content
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 22a
Chapter 25, Problem 22a

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?

Verified step by step guidance
1
Step 1: Understand the relationship between electric field strength and potential difference. The formula to use is V = E × d, where V is the potential difference, E is the electric field strength, and d is the separation distance between the plates.
Step 2: Convert the given values into consistent units. The spacing between the plates is given as 2.0 mm, which should be converted to meters: d = 2.0 mm = 2.0 × 10^-3 m.
Step 3: Substitute the given values into the formula. The electric field strength is E = 1.0 × 10^5 V/m, and the separation distance is d = 2.0 × 10^-3 m.
Step 4: Multiply the electric field strength (E) by the separation distance (d) to calculate the potential difference (V). The formula becomes V = (1.0 × 10^5 V/m) × (2.0 × 10^-3 m).
Step 5: The result of the multiplication will give the potential difference across the capacitor. Ensure the units are consistent and verify the calculation process.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

Key Concepts

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

Electric Field (E)

The electric field (E) is a vector field that represents the force per unit charge experienced by a positive test charge placed in the field. It is measured in volts per meter (V/m) and indicates how strong the electric force is in a given region. In the context of a parallel-plate capacitor, the electric field is uniform between the plates and is directly related to the potential difference and the distance between the plates.
Recommended video:
Guided course
03:16
Intro to Electric Fields

Potential Difference (V)

Potential difference, or voltage (V), is the work done per unit charge to move a charge between two points in an electric field. It is measured in volts (V) and is a crucial concept in understanding how capacitors store energy. The potential difference across a capacitor can be calculated using the formula V = E * d, where E is the electric field strength and d is the distance between the plates.
Recommended video:
Guided course
07:04
Potential Difference Between Two Charges

Capacitance (C)

Capacitance (C) is the ability of a capacitor to store charge per unit voltage and is measured in farads (F). It depends on the physical characteristics of the capacitor, such as the area of the plates and the distance between them, as well as the dielectric material used. While not directly asked in this question, understanding capacitance is essential for grasping how capacitors function and their role in electric circuits.
Recommended video:
Guided course
08:02
Capacitors & Capacitance (Intro)
Related Practice
Textbook Question

A 250 pg dust particle has charge −250e. Its speed is 2.0 m/s at point 1, where the electric potential is V₁=2000 V. What speed will it have at point 2, where the potential is V₂=−5000 V? Ignore air resistance and gravity.

1
views
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. How much charge is on each plate?

1
views
Textbook Question

A student wants to make a very small particle accelerator using a 9.0 V battery. What speed will an electron have after being accelerated from rest through the 9.0 V potential difference?

2
views
Textbook Question

In proton-beam therapy, a high-energy beam of protons is fired at a tumor. As the protons stop in the tumor, their kinetic energy breaks apart the tumor's DNA, thus killing the tumor cells. For one patient, it is desired to deposit 0.10 J of proton energy in the tumor. To create the proton beam, protons are accelerated from rest through a 10,000 kV potential difference. What is the total charge of the protons that must be fired at the tumor?

1
views
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?

1
views
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?