Chapter 23: Q70P (page 684)

Charge of uniform volume density $ρ = 3.2 μC / m^{3}$ fills a nonconducting solid sphere of radius $5.0 cm .$ . What is the magnitude of the electric field
(a) $3.5 cm$ and
(b) $8.0 cm$ from the sphere’s center?

Short Answer

Expert verified

a) The magnitude of the electric fieldat3.5 cm from the center of the sphereis ${4.2×10}^{3} N / C$ .

b) The magnitude of the electric field at 8.0 cm from the center of the sphere is $2.4 \times 10^{3} N / C$ .

Step by step solution

Listing the given quantities

Volume Charge density $ρ = 3.2 μC / m^{3}$
radius of the sphere = 5.0 cm

Understanding the concept of charge density and electric field

Since the charge distribution is uniform, we can find the total charge q by multiplying by the spherical volume $(\frac{4}{3} {πr}^{3})$ with r = R = 0.050 m.

This gives q = 1.68 nC.

Step 3: (a) Calculations for the magnitude of the electric field at r = 3.5 cm from the center of the sphere

Given r = 0.035 m.

The electric field can be calculated as:

$E_{internal} = \frac{|q| r}{4 {πε}_{0} R^{3}} = \frac{(9.0 \times 10^{9} N . m^{2} / C^{2}) (1.68 \times 10^{- 9} C) (0.035 m)}{{(0.050 m)}^{3}} = 4.2 \times 10^{3} N / C$

The magnitude of the electric field at 3.5 cm from the center of the sphere is $4.2 \times 10^{3} N / C$ .

(b) Calculations for the magnitude of the electric field at r=8.0 cm from the center of the sphere

Outside the sphere, we have (with r = 0.080 m), then the electric field can be calculated as:

$E_{internal} = \frac{|q| r}{4 {πε}_{0} R^{2}} = \frac{(9.0 \times 10^{9} N . m^{2} / C^{2}) (1.68 \times 10^{- 9} C)}{{(0.080 m)}^{2}} = 2.4 \times 10^{3} N / C$

The magnitude of the electric field at 8.0 cm from the center of the sphere is $2.4 \times 10^{3} N / C$ .

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Charge of uniform volume density $ρ = 3.2 μC / m^{3}$ fills a nonconducting solid sphere of radius $5.0 cm .$ . What is the magnitude of the electric field
(a) $3.5 cm$ and
(b) $8.0 cm$ from the sphere’s center?

Short Answer

Step by step solution

Listing the given quantities

Understanding the concept of charge density and electric field

Step 3: (a) Calculations for the magnitude of the electric field at r = 3.5 cm from the center of the sphere

(b) Calculations for the magnitude of the electric field at r=8.0 cm from the center of the sphere

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Most popular questions from this chapter

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Charge of uniform volume density ρ=3.2 μC/m3 fills a nonconducting solid sphere of radius5.0cm. . What is the magnitude of the electric field(a) 3.5cmand(b) 8.0cmfrom the sphere’s center?

Short Answer

Step by step solution

Listing the given quantities

Understanding the concept of charge density and electric field

Step 3: (a) Calculations for the magnitude of the electric field at r = 3.5 cm from the center of the sphere

(b) Calculations for the magnitude of the electric field at r=8.0 cm from the center of the sphere

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Modern Physics

Quantum Physics

Nuclear Physics

Fundamentals of Physics

Electromagnetism

Electrostatics

Study anywhere. Anytime. Across all devices.

Charge of uniform volume density $ρ = 3.2 μC / m^{3}$ fills a nonconducting solid sphere of radius $5.0 cm .$ . What is the magnitude of the electric field
(a) $3.5 cm$ and
(b) $8.0 cm$ from the sphere’s center?