Two very thin conducting, concentric spheres have radii R1 and R2 as shown in the figure.

Sunday, 26 August 2018

12:53 AM

Machine generated alternative text: Question: b Figure 1 _ Two very thin conducting, concentric spheres nave ditterent radii. The inner Snell with radius RI nas charge A-qi and the outer Snell witn radius R2 has charge +q2. Point a lies outside the two shells ra > R2 and point b lies beüeen the shells RI < "b < 112. Two very thin conducting, concentric spheres nave radii RI and R2 as shown in the figure. The inner sphere (with radius RI) has a total charge qi which distributes itself uniformly over the surface. The outer sphere (witn radius R2) has a total charge q2 which spreads itself uniformly over the surtace. Define the potential V an infinite distance trom a point charge as 0, (V = 0). When giving expressions use co, not Coulomb's constant part 1) Write an expression tor Va, the voltage at point a, a distance ru trom the centre of the spheres. (ru > R2. ) part 2) Write an expression tor Vb, the voltage at point b a distance Tb from the centre ottne spheres. (R2 < rt, < RI). part 3) Write an expression tor Vo, the voltage at the centre ottne sphere. Check

 

Machine generated alternative text: Feedback tor Part 1) using Gauss's law it can be shown that the electric field a distance r trom the centre of the spheres is 4m•2 EO perpendicularly out trom the spheres. Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 2) -fi.ds (q1++2) 47rE0 (q1+q2) 4KE0 (q1++2) 47rE0 47rEor„ • dr 4TT2E0 dr r2 This question should be done in two steps, we should calculate the potential between eo and R2 (when E is given by ) and then between R2 and rb where E is given by , this can be obtained using Gauss's law. v.R2 = ( ra in previous part is R2.) 2 _gx_ dr R'2 4TE0r2 47rE0 JR2 r2 4TE0 1 47rE0 rb r R2 1 vtotal + 47rE0 R2 47rE0 R2 47rE0 rb • Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11

 

Machine generated alternative text: Feedback tor Part 3) As witn the previous part this question should be done in parts, witn different electric fields in each part 47rE0 R2 • Then from R2 —+ RI Finally trom RI —+ 0. Inside the sphere, E V RYRI = 47rE0 0, so: Vtotal 47rE0 R 2 47rE0 R2 47rE0 RI • Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 A correct answer is , which can be typed in as follows: A correct answer is , which can be typed in as follows: A correct answer is , which can be typed in as follows:

 

 

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