{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

06 - Solutions

# 06 - Solutions - Solutions for Homework 6 Problems from...

This preview shows pages 1–2. Sign up to view the full content.

Solutions for Homework 6 Problems from Chapter 23 13. ( a ) The electric field at the surface of the Earth is the same as that of a point charge, E = Q 4 !" 0 r 0 2 . The electric potential at the surface, relative to V ( ! ) = 0 is given by Eq. 23-5. Writing this in terms of the electric field and radius of the earth gives the electric potential. V = Q 4 !" 0 r 0 = Er 0 = # 150V m ( ) 6.38 \$ 10 6 m ( ) = # 0.96 GV ( b ) Part ( a ) demonstrated that the potential at the surface of the earth is 0.96 GV lower than the potential at infinity. Therefore if the potential at the surface of the Earth is taken to be zero, the potential at infinity must be V ( ! ) = 0.96 GV . If the charge of the ionosphere is included in the calculation, the electric field outside the ionosphere is basically zero. The electric field between the earth and the ionosphere would remain the same. The electric potential, which would be the integral of the electric field from infinity to the surface of the earth, would reduce to the integral of the electric field from the ionosphere to the earth. This would result in a negative potential, but of a smaller magnitude. 21. We first need to find the electric field. Since the charge distribution is spherically symmetric, Gauss’s law tells us the electric field everywhere. r E ° d r A ° ! = E 4 " r 2 ( ) = Q encl # 0 \$ E = 1 4 "# 0 Q encl r 2 If r < r 0 , calculate the charge enclosed in the manner of Example 22-5. Q encl = ! E dV " = ! 0 1 # r 2 r 0 2 \$ % & ( ) 4 * r 2 dr 0 r " = 4 *! 0 r 2 # r 4 r 0 2 \$ % & ( ) dr 0 r " = 4 *! 0 r 3 3 # r 5 5 r 0 2 \$ % & ( ) The total charge in the sphere is the above expression evaluated at r = r 0 . Q total = 4 !" 0 r 0 3 3 # r 0 5 5 r 0 2 \$ % & ( ) = 8 !" 0 r 0 3 15 Outside the sphere, we may treat it as a point charge, and so the potential at the surface of the sphere is given by Eq. 23-5, evaluated at the surface of the sphere. V r = r 0 ( ) = 1 4 !" 0 Q total r 0 = 1 4 !" 0 8 !# 0 r 0 3 15 r 0 = 2 # 0 r 0 2 15 " 0 The potential inside is found from Eq. 23-4a. We need the field inside the sphere to use Eq.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern