A thin rod of length 2ℓ is centered on the x axis as shown in Fig. 23–46. The rod carries a uniformly distributed charge Q. Determine the potential V as a function of y for points along the y axis. Let V = 0 at infinity.
Ch. 23 - Electric Potential
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook
Chapter 22, Problem 19
Suppose the end of your finger is charged.
(a) Estimate the breakdown voltage in air for your finger.
(b) About what surface charge density would have to be on your finger at this voltage?
Verified step by step guidance1
Estimate the breakdown voltage in air for your finger by using the approximate electric field strength required for air to break down, which is about 3 × 10^6 V/m. Assume the distance between your finger and the object is small, say 1 mm (0.001 m). Use the formula for electric field: , where E is the electric field, V is the voltage, and d is the distance.
Rearrange the formula to solve for the breakdown voltage . Substitute the values for E (3 × 10^6 V/m) and d (0.001 m) to estimate the breakdown voltage.
To calculate the surface charge density, use the relationship between the electric field and surface charge density: , where σ is the surface charge density and ε 0 is the permittivity of free space (8.85 × 10^-12 C^2/N·m^2).
Rearrange the formula to solve for the surface charge density: . Substitute the value of E (3 × 10^6 V/m) and ε 0 (8.85 × 10^-12 C^2/N·m^2) to find the surface charge density.
Combine the results from parts (a) and (b) to understand the relationship between the breakdown voltage and the surface charge density on your finger. This will give you a complete picture of the physical situation.

Verified video answer for a similar problem:
This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
6mWas this helpful?
Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Breakdown Voltage
Breakdown voltage is the minimum voltage that causes a portion of an insulator to become electrically conductive. In air, this occurs when the electric field strength exceeds approximately 3 million volts per meter (3 MV/m). This phenomenon is crucial for understanding how and when air can conduct electricity, such as during lightning strikes or electrical discharges.
Recommended video:
Guided course
Dielectric Breakdown
Surface Charge Density
Surface charge density refers to the amount of electric charge per unit area on a surface. It is typically measured in coulombs per square meter (C/m²). To estimate the surface charge density on a charged object, one can use the relationship between the electric field and the charge distribution, which is essential for determining how much charge is needed to reach the breakdown voltage in air.
Recommended video:
Guided course
Surface Charge Density
Electric Field
An electric field is a region around a charged object where other charged objects experience a force. The strength of the electric field (E) is defined as the force (F) per unit charge (q) and is measured in volts per meter (V/m). Understanding electric fields is vital for analyzing how charges interact and for calculating the breakdown voltage and surface charge density in the context of the question.
Recommended video:
Guided course
Intro to Electric Fields
Related Practice
Textbook Question
1
views
Textbook Question
What minimum radius must a large conducting sphere (of an electrostatic generating machine) have if it is to be at 45,000 V without discharge into the air? How much charge will it carry?
2
views
Textbook Question
A very long conducting cylinder (length ℓ) of radius R₀ (R₀ ≪ ℓ) carries a uniform surface charge density σ (C/m²). The cylinder is at an electric potential V₀. Determine the potential, at points far from the end, at a distance R from the center of the cylinder for
(a) R > R₀
(b) R < R₀.
(c) Is V = 0 at R = ∞ (assume ℓ = ∞ )? Explain. [Hint: Recall Gauss’s law.]
2
views
Textbook Question
Two point charges, 3.4 μC and -2.0 μC, are placed 8.0 cm apart on the x axis. At what points along the x axis are
(a) the electric field zero and
(b) the potential zero? Let V = 0 at r = ∞.
2
views
