Ch. 23 - Electric Potential

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 23 - Electric Potential

Problem 44

Chapter 22, Problem 44

A metal sphere of radius r₀ = 0.35 m carries a charge Q = 0.50 μC. Equipotential surfaces are to be drawn for 100-V intervals outside the sphere. Determine the radius r of the first equipotential from the surface.

Verified step by step guidance

Understand the concept: The problem involves equipotential surfaces around a charged sphere. For a charged sphere, the electric potential at a distance r from the center is given by the formula:

V = \frac{k Q}{r}

, where

k

is Coulomb's constant,

Q

is the charge, and

r

is the distance from the center of the sphere.

Set up the equation for the potential: The potential at the surface of the sphere (radius

r_{0}

) is given by

V_{0} = \frac{k Q}{r_{0}}

. For the first equipotential surface, the potential decreases by 100 V, so the new potential is

V = V_{0} - 100

Rearrange the formula for the radius: Substitute the new potential into the formula for electric potential and solve for

r

. The equation becomes

r = \frac{k Q}{V}

, where

V

is the new potential.

Substitute the known values: Use the given values for

k = 8.99 \times 10^{9}

N·m²/C²,

Q = 0.50 \times 10^{-6}

C, and the new potential

V = V_{0} - 100

to calculate the radius

r

Interpret the result: The calculated radius

r

represents the distance from the center of the sphere to the first equipotential surface where the potential has decreased by 100 V. This process can be repeated for subsequent equipotential surfaces by further decreasing the potential in 100-V intervals.

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Key Concepts

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

Electric Potential

Electric potential, or voltage, is the amount of electric potential energy per unit charge at a point in an electric field. It is measured in volts (V) and indicates how much work would be done to move a charge from a reference point to a specific point in the field. For a charged sphere, the potential outside the sphere can be calculated using the formula V = k * Q / r, where k is Coulomb's constant.

Equipotential Surfaces

Equipotential surfaces are three-dimensional surfaces where the electric potential is constant. No work is required to move a charge along an equipotential surface, as the potential difference is zero. In the context of a charged sphere, these surfaces are concentric spheres centered on the charge, with the potential increasing as one moves away from the sphere.

Coulomb's Law

Coulomb's Law describes the force between two point charges. It states that the force (F) between two charges is directly proportional to the product of the magnitudes of the charges (Q₁ and Q₂) and inversely proportional to the square of the distance (r) between them. This law is fundamental in understanding how electric fields and potentials are generated by charged objects, such as the metal sphere in this problem.

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Coulomb's Law

Related Practice

Textbook Question

A dust particle with mass of 0.050 g and a charge of 2.0 x 10⁻⁶ C is in a region of space where the potential is given by V(x) = (2.0 V/m²) x² - (3.0 V/m³)x³. If the particle starts at x = 2.5m, what is the initial acceleration of the charge?

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Textbook Question

Calculate the electric potential due to a tiny dipole whose dipole moment is 4.8 x 10⁻³⁰ Cm at a point 4.1 x 10⁻⁹ m away if this point is along the axis of the dipole nearer the positive charge.

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Textbook Question

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.

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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.

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Textbook Question

Calculate the electric potential due to a tiny dipole whose dipole moment is 4.8 x 10⁻³⁰ Cm at a point 4.1 x 10⁻⁹ m away if this point is 45° above the axis but nearer the positive charge.

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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 = ∞.

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