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Ch 22: Electric Charges and Forces
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 22: Electric Charges and ForcesProblem 34b
Chapter 22, Problem 34b

A 0.10 g honeybee acquires a charge of +23 pC while flying. What electric field (strength and direction) would allow the bee to hang suspended in the air?

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Step 1: Begin by identifying the forces acting on the honeybee. The honeybee is suspended in the air, meaning the electric force acting on it must balance the gravitational force. The gravitational force can be calculated using the formula: Fg=mg, where m is the mass of the honeybee and g is the acceleration due to gravity (approximately 9.8 m/s²).
Step 2: Calculate the gravitational force acting on the honeybee. Convert the mass of the honeybee from grams to kilograms (0.10 g = 0.0001 kg) and substitute it into the formula for gravitational force: Fg=0.0001×9.8. This gives the magnitude of the gravitational force.
Step 3: Recall the formula for electric force: Fe=qE, where q is the charge on the honeybee and E is the electric field strength. To suspend the honeybee, the electric force must equal the gravitational force: Fe=Fg. Substitute the expression for Fe into this equation: qE=mg.
Step 4: Solve for the electric field strength E. Rearrange the equation to isolate E: E=mg/q. Substitute the values for m (0.0001 kg), g (9.8 m/s²), and q (+23 pC = 23 × 10⁻¹² C).
Step 5: Determine the direction of the electric field. Since the honeybee has a positive charge, the electric field must point upward to exert an upward force on the honeybee, counteracting the downward gravitational force. This ensures the honeybee remains suspended in the air.

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Key Concepts

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

Electric Field

An electric field is a region around a charged particle where other charged particles experience a force. It is represented by the symbol E and is measured in newtons per coulomb (N/C). The direction of the electric field is defined as the direction a positive test charge would move in the field. Understanding electric fields is crucial for analyzing the forces acting on charged objects, such as the honeybee in this scenario.
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Force on a Charged Particle

The force experienced by a charged particle in an electric field is given by Coulomb's law, expressed as F = qE, where F is the force, q is the charge of the particle, and E is the electric field strength. This relationship indicates that the force is directly proportional to both the charge and the strength of the electric field. In the case of the honeybee, this force must counteract its weight to achieve suspension in the air.
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Weight of the Honeybee

The weight of an object is the force exerted on it due to gravity, calculated as W = mg, where m is the mass and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth). For the honeybee, with a mass of 0.10 g (or 0.0001 kg), its weight must be determined to find the necessary electric field that will balance this force, allowing the bee to remain suspended in the air.
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Related Practice
Textbook Question

Two 1.0 g spheres are charged equally and placed 2.0 cm apart. When released, they begin to accelerate at 150 m/s2. What is the magnitude of the charge on each sphere?

Textbook Question

A 3.00-cm-long spring has a small plastic bead glued to each end. Charging each bead to −25 nC expands the spring by 0.50 cm. What is the value of the spring constant?

Textbook Question

The electric field at a point in space is E=(400i^+100j^)E=(400\(\hat{i}\)+100\(\hat{j}\)) N/C. What is the electric force on a proton at this point? Give your answer in component form.

Textbook Question

The electric field at a point in space is E=(400i^+100j^)E=(400\(\hat{i}\)+100\(\hat{j}\)) N/C. What is the magnitude of the electron’s acceleration?

Textbook Question

Objects A and B are both positively charged. Both have a mass of 100 g, but A has twice the charge of B. When A and B are placed 10 cm apart, B experiences an electric force of 0.45 N. What is the charge on A?

Textbook Question

A −12nC-12\,\(\text{nC}\) charge is located at (x,y)=(1.0cm,0cm)(x, y)=(1.0\,\(\text{cm}\), 0\,\(\text{cm}\)). What are the electric fields at the positions (x,y)=(5.0cm,0cm),(−5.0cm,0cm)(x,y)=(5.0\,\(\text{cm}\),0\,\(\text{cm}\)),(-5.0\,\(\text{cm}\),0\,\(\text{cm)}\), and (0cm,5.0cm)(0\,\(\text{cm}\), 5.0\,\(\text{cm}\))? Write each electric field vector in component form.

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