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The charged balloon in FIGURE Q24.7 expands as it is blown up, increasing in size from the initial to final diameters shown. Do the electric field strengths at points 1, 2, and 3 increase, decrease, or stay the same? Explain your reasoning for each.

Short Answer

Expert verified

Electric field at point 1 is the same

Electric field at point 2 decreases

Electric field at point 3 is the same

Step by step solution

01

Given information and theory used

Given Figure :

Theory used :

At all sites in electrostatic equilibrium inside a conductor, the Electric Field is zero. However, all surplus charges on the conductor accumulate on the outside surface, and as further charges are added, they spread out on the outer surface until they reach the electrostatic equilibrium points. The electric field at the surface of a charged conductor in the form

Esurface=0, where is the surface charge density.

02

Determining if the electric field strengths at points 1, 2, and 3 increase, decrease, or stay the same 

The surface charge density is not constant and is a physical parameter that relies on the conductor's geometry. Calculating the surface charge density using the electric field outside the conductor :

Because the electric field at point 1 is the same at both the beginning and the end of the balloon, the electric field at point 1 is the same.

The electric field at point 2 is weaker. Because point 2 is outside the balloon at first, an electric field exists due to charges on the surface; however, as the balloon expands, point 2 becomes inside the balloon, and the electric field vanishes. The electric field shrinks in this situation.

The electric field at point 3 is same. Point 3 lies outside the balloon both before and after the expansion, therefore the electric field is the same.

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Most popular questions from this chapter

Newton鈥檚 law of gravity and Coulomb鈥檚 law are both inversesquare laws. Consequently, there should be a 鈥淕auss鈥檚 law for gravity.鈥 a. The electric field was defined as E u = F u on q /q, and we used this to find the electric field of a point charge. Using analogous reasoning, what is the gravitational field g u of a point mass?

Write your answer using the unit vector nr, but be careful with signs; the gravitational force between two 鈥渓ike masses鈥 is attractive, not repulsive. b. What is Gauss鈥檚 law for gravity, the gravitational equivalent of Equation 24.18? Use 桅G for the gravitational flux, g u for the gravitational field, and Min for the enclosed mass. c. A spherical planet is discovered with mass M, radius R, and a mass density that varies with radius as r = r011 - r/2R2, where r0 is the density at the center. Determine r0 in terms of M and R. Hint: Divide the planet into infinitesimal shells of thickness dr, then sum (i.e., integrate) their masses. d. Find an expression for the gravitational field strength inside the planet at distance r 6 R.

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An infinite cylinder of radius Rhas a linear charge density . The volume charge density C/m3within the cylinder (rR)is (r)=r0/R, where 0is a constant to be determined.

a. Draw a graph of versus localid="1648911863544" xfor an x-axis that crosses the cylinder perpendicular to the cylinder axis. Let xrange from 2Rto 2R.

b. The charge within a small volume dVis dq=dV. The integral of dVover a cylinder of length localid="1648848405768" Lis the total charge Q=Lwithin the cylinder. Use this fact to show that 0=3/2R2.

Hint: Let dVbe a cylindrical shell of length L, radius r, and thickness dr. What is the volume of such a shell?

c. Use Gauss's law to find an expression for the electric field strength Einside the cylinder, localid="1648889098349" rR, in terms of and R.

d. Does your expression have the expected value at the surface, localid="1648889146353" r=R? Explain.

FIGUREP24.38shows a solid metal sphere at the center of a hollow metal sphere. What is the total charge on (a) the exterior of the inner sphere, (b) the inside surface of the hollow sphere, and (c) the exterior surface of the hollow sphere?

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