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
Chapter 25, Problem 55

A 2.0-mm-diameter glass bead is positively charged. The potential difference between a point 2.0 mm from the bead and a point 4.0 mm from the bead is 500 V. What is the charge on the bead?

Verified step by step guidance
1
Understand the problem: The potential difference (ΔV) between two points at distances r₁ = 2.0 mm and r₂ = 4.0 mm from the center of a charged sphere (the glass bead) is given as 500 V. We need to find the charge (q) on the bead. Assume the bead behaves like a point charge, and use the formula for electric potential due to a point charge.
Recall the formula for the electric potential due to a point charge: V = (1 / (4πϵ₀)) * (q / r), where ϵ₀ is the permittivity of free space, q is the charge, and r is the distance from the charge. The potential difference between two points is given by ΔV = V₁ - V₂.
Substitute the expressions for V₁ and V₂ into the potential difference formula: ΔV = (1 / (4πϵ₀)) * (q / r₁) - (1 / (4πϵ₀)) * (q / r₂). Factor out the common terms to simplify: ΔV = (q / (4πϵ₀)) * (1 / r₁ - 1 / r₂).
Rearrange the equation to solve for q: q = ΔV * (4πϵ₀) / (1 / r₁ - 1 / r₂). Substitute the given values: ΔV = 500 V, r₁ = 2.0 mm = 0.002 m, r₂ = 4.0 mm = 0.004 m, and ϵ₀ = 8.85 × 10⁻¹² C²/(N·m²).
Perform the calculations step by step: First, calculate (1 / r₁ - 1 / r₂). Then, multiply this result by (4πϵ₀). Finally, multiply by ΔV to find q. Ensure all units are consistent throughout the calculation.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

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

Electric Potential and Potential Difference

Electric potential is the amount of electric potential energy per unit charge at a point in an electric field. The potential difference between two points is the work done to move a unit charge from one point to another. In this question, the potential difference of 500 V indicates how much energy is required to move a charge between the specified distances from the charged bead.
Recommended video:
Guided course
07:33
Electric Potential

Coulomb's Law

Coulomb's Law describes the force between two charged objects. It states that the force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This law is essential for calculating the charge on the bead based on the electric potential created at different distances.
Recommended video:
Guided course
09:52
Coulomb's Law

Relation Between Charge, Potential, and Distance

The relationship between charge, electric potential, and distance can be expressed through the formula V = kQ/r, where V is the electric potential, k is Coulomb's constant, Q is the charge, and r is the distance from the charge. This relationship allows us to determine the charge on the bead by using the given potential difference and distances from the bead.
Recommended video:
Guided course
05:45
Potential Difference Between Two Charges
Related Practice
Textbook Question

Three electrons form an equilateral triangle 1.0 nm on each side. A proton is at the center of the triangle. What is the potential energy of this group of charges?

176
views
Textbook Question

The electric potential in a region of space is given by V=V₀[(x²+2y²)/(0.10 m)²], where V₀ is a constant. A proton released from rest at (x, y)=(20 cm, 0 cm) reaches the origin with a speed of 7.5×105 m/s. At what value of y on the y-axis should a He+ ion (charge +e, mass 4 u) be released from rest to reach the origin with the same speed?

1584
views
Textbook Question

A proton is fired from far away toward the nucleus of an iron atom. Iron is element number 26, and the diameter of the nucleus is 9.0 fm. What initial speed does the proton need to just reach the surface of the nucleus? Assume the nucleus remains at rest.

110
views
Textbook Question

A 2.0-cm-diameter copper ring has 5.0×109 excess electrons. A proton is released from rest on the axis of the ring, 5.0 cm from its center. What is the proton's speed as it passes through the center of the ring?

134
views
Textbook Question

In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons, thus having mass m=4 u and charge q=2e. Suppose a uranium nucleus with 92 protons decays into thorium, with 90 protons, and an alpha particle. The alpha particle is initially at rest at the surface of the thorium nucleus, which is 15 fm in diameter. What is the speed of the alpha particle when it is detected in the laboratory? Assume the thorium nucleus remains at rest.

112
views
Textbook Question

What is the escape speed of an electron launched from the surface of a 1.0-cm-diameter glass sphere that has been charged to 10 nC?

1882
views