The total charge inside our gaussian surface is $q_{in} = lambda L$. Applying Gauss's law and applying conditions 1 and 2, we find the following expression for the curved surface. $Phi_E = oint vec{E} cdot dvec{A} = E oint dA = EA = frac{lambda L}{epsilon_0} = frac{lambda L}{epsilon_0}$ The area of the curved surface is $A = 2pi rL$. Therefore we can find an expression for E. (Use the following as necessary: $k_e$, $lambda$, and r.) $E(2pi rL) = frac{lambda L}{epsilon_0}$ $E = frac{lambda}{2pi epsilon_0 r} = frac{2k_e lambda}{r}$ (Equation 24.7) Therefore, we see that the electric field of a cylindrically symmetric charge distribution varies as 1 / r, whereas the field external to a spherically symmetric charge distribution varies as 1 / $r^2$. Equation 24.7 was also derived by integration of the field of a point charge. If the line charge in this example were of finite length, the result for E is not that given by Equation 24.7. A finite line charge does not possess sufficient symmetry to use Gauss's law because the magnitude of the electric field is no longer constant over the surface of the gaussian cylinder; the field near the ends of the line would be different from that far from the ends. Therefore, condition 1 is not satisfied in this situation. Furthermore, E is not perpendicular to the cylindrical surface at all points; the field vectors near the ends would have a component parallel to the line. Condition 2 is not satisfied. For points close to a finite line charge and far from the ends, Equation 24.7 gives a good approximation of the value of the field. Exercise 24.7 Hints: Getting Started | I'm Stuck Consider a long cylindrical charge distribution of radius R = 15 cm with a uniform charge density of $ ho = 16 ext{ C/m}^3$. Find the electric field at a distance r = 25 cm from the axis. E = 11.52e+10 N/C It might be helpful to carefully follow through the example to make sure you understand the solution. Be careful how you calculate the charge enclosed by the Gaussian surface.
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Sri K.
We want to find the electric field near an infinite charged conducting plane. Gauss' Law works for any closed surface whatsoever. The art of using it is to match the shape of the Gaussian surface to the symmetry of the problem. The figure shows a Gaussian surface of the variety called a "pillbox," a cylinder with an arbitrary area A on each end and extending for a height h from the surface. The conductor bears a surface charge density of Ģ in [C/mĀ²]. The left end of the pillbox is buried inside the conductor. By symmetry, the electric field E must be perpendicular to the surface (read that again, and look at the diagram). Always start by writing down Gauss' Law: ā® E ā dA = q_enclosed / Īµ_0 The left hand side (lhs) is an integral over the closed surface of the pillbox. Break this into the sum of three integrals: one each for the left end, the right end, and the sides of the cylinder: ā® total = ā« left end + ā« right end + ā« sides where only the first is a closed integral. These three pieces make up the closed surface. Now evaluate each of these integrals separately: Left end: ā« E ā dA = 0 (easy because E = 0 inside a conductor). Right end: ā« E ā dA = (what is the angle between E and A on the right end cap?). Sides: ā« E ā dA = (what is the angle between E and A on the right end cap? 90Ā°, right?). Sum these to get: ā® E ā dA = Now for the right hand side (rhs). You need the charge enclosed. But all of the charge resides on the surface of the conductor. What is the area of the surface inside the pillbox? How do you get from area A in [mĀ²] to charge Q in [C] using the information you have? (Hint: it's in the first paragraph.) Q_enclosed = Now set the lhs equal to the rhs (divided by Īµ_0), simplify, and you are done. Put the answer here: EA = therefore E = Notice a funny thing about this formula. It does not involve the distance from the surface. Why do you suppose this is? What assumption was made that leads to this, and when will it break down?
Madhur L.
(A) Calculate the magnitude of the electric field at a point outside the sphere. Gaussian sphere Conceptualize: Note how this problem differs from our previous discussion of Gauss's law. Now we are considering the electric field due to a distribution of charge. We found the field for various distributions of charge in the chapter entitled Electric Fields by integrating over the distribution. In this chapter, we find the electric field using Gauss's law. A uniformly charged insulating sphere of radius a and total charge Q. (a) For points outside the sphere, a large, spherical gaussian surface is drawn concentric with the sphere. In diagrams such as this one, the dotted line represents the intersection of the gaussian surface with the plane of the page. (b) For points inside the sphere, a spherical gaussian surface smaller than the sphere is drawn. Categorize: Because the charge is distributed uniformly throughout the sphere, the charge distribution has spherical symmetry and we can apply Gauss's law to find the electric field. Analyze: To reflect the spherical symmetry, let's choose a spherical gaussian surface of radius r, concentric with the sphere, as shown in Figure (a). For this choice, E and dA are parallel everywhere on the surface and E · dA = EdA. Replace E · dA in Gauss's Law with EdA: Ī¦_E = ā® E · dA = ā® EdA = Q/Īµ_0 By symmetry, E is constant everywhere on the surface, so we can remove E from the integral: ā® EdA = E ā® dA = E(4Ļr^2) = Q/Īµ_0 Solve for E. (1) E = Q/(4ĻĪµ_0r^2) (Use the following as necessary: k_e, Q, and r.) E = k_e Q / r^2 (for r > a)
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