Volume charge density is located in free space as \(\rho_{v}=2 e^{-1000 r} \mathrm{nC} / \mathrm{m}^{3}\) for \(0

Short Answer

Expert verified
Answer: The total charge enclosed by the spherical surface of radius 1mm is approximately 7.84 pC, and the electric flux density on the surface is approximately 140 N/C.

Step by step solution

01

Calculate Total Charge

To find the total charge enclosed by a sphere of radius r = 1mm, let's first define the given volume charge density: Ρv(r) = 2e^{-1000r} nC/m³ (for 0 < r < 1mm) Now, we need to integrate this charge density over the whole volume of the sphere: Total charge (Q) = ∫∫∫Ρv(r) dv Where the limits of integration are from r = 0 to r = 1mm, and we need to convert the radius into meters (0 to 0.001m). And dv = r²sinθ dr dθ dϕ where θ is from 0 to π, and ϕ is from 0 to 2π. Q = ∫(0 to 0.001) ∫(0 to π) ∫(0 to 2π) 2e^{-1000r} r² sinθ dr dθ dϕ
02

Perform the Integration

Let's solve the volume integral one variable at a time. First, we perform the integral with respect to ϕ: Q = ∫(0 to 0.001) ∫(0 to π) [2π * 2e^{-1000r} r² sinθ] dr dθ Now, we perform the integral with respect to θ: Q = ∫(0 to 0.001) [2π * -2e^{-1000r} r² (cosθ from 0 to π)] dr Q = ∫(0 to 0.001) [4π * 2e^{-1000r} r²] dr Finally, we perform the integral with respect to r: Q = -4π * 2 * 10^{-6} [e^{-1000r} r²/(-1000) + 2e^{-1000r} r/(-1000)^2 + 2e^{-1000r}/(-1000)^3] from r=0 to r=(1mm = 0.001m) By plugging r = 0.001m and r = 0 into the above expression and calculating we get, Q ≈ 7.84 * 10^{-12} C The total charge enclosed by the sphere is approximately 7.84 pC.
03

Calculate Electric Flux Density using Gauss's Law

Now that we have the total enclosed charge, we can use Gauss's Law to find the electric flux density at r=1mm. Gauss's Law states that the total electric flux through a closed surface is equal to the total enclosed charge divided by the permittivity of free space (ε0). Flux = Q / ε0 The electric flux density on the surface (Dr) is equal to the total electric flux divided by the surface area of the sphere (A). Surface area of the sphere A = 4πr². In this case, r = 1mm = 0.001m. Dr = Flux / A = (Q / ε0) / (4πr²) By plugging the known values into the equation: ε0 = 8.85 * 10^{-12} C²/Nm² Q ≈ 7.84 * 10^{-12} C r = 0.001m Dr ≈ ((7.84 * 10^{-12} C) / (8.85 * 10^{-12} C²/Nm²)) / (4π(0.001m)²) Dr ≈ 140 N/C The electric flux density on the surface r = 1mm is approximately 140 N/C.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A certain light-emitting diode (LED) is centered at the origin with its surface in the \(x y\) plane. At far distances, the LED appears as a point, but the glowing surface geometry produces a far-field radiation pattern that follows a raised cosine law: that is, the optical power (flux) density in watts \(/ \mathrm{m}^{2}\) is given in spherical coordinates by $$ \mathbf{P}_{d}=P_{0} \frac{\cos ^{2} \theta}{2 \pi r^{2}} \mathbf{a}_{r} \quad \text { watts } / \mathrm{m}^{2} $$ where \(\theta\) is the angle measured with respect to the direction that is normal to the LED surface (in this case, the \(z\) axis), and \(r\) is the radial distance from the origin at which the power is detected. \((a)\) In terms of \(P_{0}\), find the total power in watts emitted in the upper half-space by the LED; (b) Find the cone angle, \(\theta_{1}\), within which half the total power is radiated, that is, within the range \(0<\theta<\theta_{1} ;\) ( \(c\) ) An optical detector, having a \(1-\mathrm{mm}^{2}\) cross-sectional area, is positioned at \(r=1 \mathrm{~m}\) and at \(\theta=45^{\circ}\), such that it faces the \(\mathrm{LED}\). If one milliwatt is measured by the detector, what (to a very good estimate) is the value of \(P_{0}\) ?

Given the flux density \(\mathbf{D}=\frac{16}{r} \cos (2 \theta) \mathbf{a}_{\theta} \mathrm{C} / \mathrm{m}^{2}\), use two different methods to find the total charge within the region \(1

Use Gauss's law in integral form to show that an inverse distance field in spherical coordinates, \(\mathbf{D}=A a_{r} / r\), where \(A\) is a constant, requires every spherical shell of \(1 \mathrm{~m}\) thickness to contain \(4 \pi A\) coulombs of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation with \(r\).

An electric flux density is given by \(\mathbf{D}=D_{0} \mathbf{a}_{\rho}\), where \(D_{0}\) is a given constant. (a) What charge density generates this field? \((b)\) For the specified field, what total charge is contained within a cylinder of radius \(a\) and height \(b\), where the cylinder axis is the \(z\) axis?

A cube is defined by \(1

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free