Chapter 4: Q33P (page 747)

A small sphere with mass $4.00 \times 10^{- 6} kg$ and charge $5.00 \times 10^{- 8} C$ hangs from a thread near a very large, charged insulating sheet (Fig.P22.33). The charge density on the surface of the sheet is uniform and equal to $- 2.50 \times 10^{- 9} C / m^{2}$ . Find the angle of the thread.

Short Answer

Expert verified

The angle of the thread is $10.2 °$

Step by step solution

Define free body diagram and given data

Free body diagram,

Given,

$m = 4 \times 10^{- 6} kg σ = 2.50 \times 10^{- 9} C \times m^{- 2} q = 5.0 \times 10^{- 8} C$

Simplify the direction of force

The ball is in equilibrium so the net force in each direction is equal to zero so the force indirection is given by

$Tsin (θ) = F_{e} . . . . . . (1)$

Also, the force in y direction is

$Tcos (θ) = mg . . . . (2)$

Dividing (1) over (2) to get

$\tan (θ) = \frac{F_{e}}{mg} Þ (3)$

The electric force is given by

$F_{e} = E_{q} = \frac{qσ}{2 \overset{'}{o_{0}}} = \frac{2.5 \times 10^{- 9} \times 5 \times 10^{- 8}}{2 \times 8.85 \times 10^{- 12}} = 7.06 \times 10^{- 6} N$

Also, the gravitational force is given by

$mg = 4 \times 10^{- 6} \times 9.8 = 3.9 \times 10^{- 5} N$

Substitution in (3) to get

$\tan (θ) = \frac{7.06 \times 10^{- 6}}{3.9 \times 10^{- 5}} = 0.18 θ = 10.2 °$

Hence, the angle of the thread is $10.2 °$

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Short Answer

Step by step solution

Define free body diagram and given data

Simplify the direction of force

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A small sphere with mass 4.00×10-6 kgand charge 5.00×10-8Changs from a thread near a very large, charged insulating sheet (Fig.P22.33). The charge density on the surface of the sheet is uniform and equal to -2.50×10-9C/m2. Find the angle of the thread.

Short Answer

Step by step solution

Define free body diagram and given data

Simplify the direction of force

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Engineering Physics

Force

Magnetism

Mechanics and Materials

Medical Physics

Electricity

Study anywhere. Anytime. Across all devices.