Explanation 1) Concept The electric potential at a point is the amount of work done in moving a unit positive charge from infinity to that point against the electric forces. 2) Formulae i. Electric potential at a point due to a point charge q V = k q r Where k- 1 4 π ε o c a l l e d Coulomb’s constant which is equal to 8.99 × 1 0 9 N . m 2 / C 2 r- is the distance of the point from the charge ii. The electric potential difference between two points A and B is V B - V A = - ∫ A B E d s iii. Gauss law says Electric flux through a surface ϕ = ∮ E ⃑ . d A ⃑ = q e n c ε 0 3) Given Charge on the thick spherical shell=Q Uniform volume charge density= ρ V=0 at infinity 4) Calculations (a) For r > r 2 the field is like that of a point charge and Potential, V = k Q r where the zero of potential was taken to be at infinity. For r > r 2 , V = k Q r (b) For r 2 > r > r 1, Volume charge density ρ = Q V -----------------------(1) Volume of the sphere, V = 4 π r 2 3 - r 1 3 3 --------------------(2) Using (2) in (1 ) gives us ρ = 3 Q 4 π r 2 3 - r 1 3 --------------------(3) Charge enclosed by the Gaussian surface is, q enc = 4 π r 3 - r 1 3 ρ 3 Using (3) in the above expression, we get q enc = Q r 3 - r 1 3 r 2 3 - r 1 3 According to Gauss law 4 π ε o r 2 E= Q r 3 - r 1 3 r 2 3 - r 1 3 On rearranging for E E= Q r 3 - r 1 3 4 π ε o r 2 r 2 3 - r 1 3 If V s is the electric potential at the outer surface of the shell, then the potential a distance r from the center is given by V = V s - ∫ r 2 r E d r = V s - Q 4 π ε 0

Fundamentals Of Physics 11th Edition Loose-leaf Print Companion Volume 2 With Wileyplus Card Set

11th Edition

ISBN: 9781119463252

Author: David Halliday

Publisher: John Wiley and Sons

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Question

Chapter 24, Problem 95P

To determine

To find:

the electric potential V as a function of distance r from the center of the distribution, considering regions

(a) r > r₂,

(b) r₂ > r > r₁, and

(c) r< r₁

(d) whether these solutions agree with each other at r= r₂ and r = r₁

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Chapter 24, Problem 94P

Chapter 24, Problem 96P

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Chapter 24 Solutions

Fundamentals Of Physics 11th Edition Loose-leaf Print Companion Volume 2 With Wileyplus Card Set

Ch. 24 - Figure 24-24 shows eight particles that form a...Ch. 24 - Figure 24-25 shows three sets of cross sections of...Ch. 24 - Figure 24-26 shows four pairs of charged...Ch. 24 - Figure 24-27 gives the electric potential V as a...Ch. 24 - Figure 24-28 shows three paths along which we can...Ch. 24 - Figure 24-29 shows four arrangement? of charged...Ch. 24 - Figure 24-30 shows a system of three charged...Ch. 24 - In the situation of Question 7, is the work done...Ch. 24 - Figure 24-26 shows four pairs of charged particles...Ch. 24 - a In Fig. 24-31a, what is the potential at point P...

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the electric potential V as a function of distance r from the center of the distribution, considering regions (a) r &gt; r 2 , (b) r 2 &gt; r &gt; r 1 , and (c) r&lt; r 1 (d) whether these solutions agree with each other at r= r 2 and r = r 1

To find: the electric potential V as a function of distance r from the center of the distribution, considering regions (a) r > r2, (b) r2 > r > r1, and (c) r< r1 (d) whether these solutions agree with each other at r= r2 and r = r1

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Chapter 24 Solutions

the electric potential V as a function of distance r from the center of the distribution, considering regions (a) r > r 2 , (b) r 2 > r > r 1 , and (c) r< r 1 (d) whether these solutions agree with each other at r= r 2 and r = r 1

To find:

the electric potential V as a function of distance r from the center of the distribution, considering regions

(a) r > r₂,

(b) r₂ > r > r₁, and

(c) r< r₁

(d) whether these solutions agree with each other at r= r₂ and r = r₁