Two solid, neutral conducting spheres of radius r are connected by a conductor with length L−2r so that the centres of the spheres are a distance L apart.

Sunday, 26 August 2018

1:06 AM

Untitled picture.png Machine generated alternative text: Question: Two solid, neutral conducting spheres ot radius r are connected by a conductor with length L 2r so that the centres ot the spheres are a distance L apart Figure 1 _ Two uncharged spheres ot radius r are a distance L apart and connected by a conductor. A charge +q is now brought close to the right hand sphere. It is a distance a trom the centre ottne sphere along the axis joining the two spheres as shown in the tigure_ a Figure 2. A charge -eq is placed a distance a from the sphere on the right and a + L from the sphere on the right. For this question assume that T a L. In this question use eo rather than Coulomb's constant in your expressions. Take the voltage an infinite distance trom a point charge to be zero, part 1) The charge +q will induce a negative charge, —Q, in the right hand sphere. What is the potential at the centre of the right hand sphere due to just this induced charge? part 2) Write an expression tor V in the centre of the right hand sphere accounting tor all the charges in the system. Vright part 3) using this expression, obtain an expression tor Q, the magnitude of the induced charge. Make sure you take appropriate limits, L > > a > > T. Check Untitled picture.png Machine generated alternative text: Feedback tor Part 1) As the sphere is a conductor all the charges will be found on the surface so —Q is all a distance r from the centre of the sphere. Thus the potential is —Q 47TEor and in this case on the sphere E q = —Q.) (as the potential due to a point charge a distance r away is -'NEO Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 2) When the —Q charge is induced on the right hand sphere an equal and opposite charge -HQ is induced on the lett hand sphere. As L r this can be treated as a point charge which contributes to the voltage at the centre of the right hand sphere. The point charge -eq contributes -Q 47rE0 a • Q 47rE0 r 4 néoa Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 3) As the two conducting spheres are connected witn a conductor the potential is the same at their centres_ using a similar method to the right hand sphere we can write: Then Vright Vieft SO: —a = LandasL L Vleft ¯ —Q Q 47rE0 L 47rE0 r Q q 47rE0 a 47rE0 L Q 47rE0 r Q 4TE0(L+a) • Q 47rE0(L+a) • Q q L4-a q L4-a a(L+a) q L 2r-2L rL x -qLa(L + a) 2(r-L) rL2 q rL2 q rq Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Ink Drawings Ink Drawings   Ink Drawings Ink Drawings Ink Drawings Ink Drawings   
Untitled picture.png Machine generated alternative text: Feedback tor Part 1) As the sphere is a conductor all the charges will be found on the surface so —Q is all a distance r from the centre of the sphere. Thus the potential is —Q 47TEor and in this case on the sphere E q = —Q.) (as the potential due to a point charge a distance r away is -'NEO Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 2) When the —Q charge is induced on the right hand sphere an equal and opposite charge -HQ is induced on the lett hand sphere. As L r this can be treated as a point charge which contributes to the voltage at the centre of the right hand sphere. The point charge -eq contributes -Q 47rE0 a • Q 47rE0 r 4 néoa Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Feedback tor Part 3) As the two conducting spheres are connected witn a conductor the potential is the same at their centres_ using a similar method to the right hand sphere we can write: Then Vright Vieft SO: —a = LandasL L Vleft ¯ —Q Q 47rE0 L 47rE0 r Q q 47rE0 a 47rE0 L Q 47rE0 r Q 4TE0(L+a) • Q 47rE0(L+a) • Q q L4-a q L4-a a(L+a) q L 2r-2L rL x -qLa(L + a) 2(r-L) rL2 q rL2 q rq Marks for this submissiom 0.00/0.33. This submission attracted a penalty of 0.11 Untitled picture.png Machine generated alternative text: A correct answer — A correct answer — , which can be typed in as follows: , which can be typed in as follows: 4mL • o ( -Q/ ) a epsi lon_e) A correct answer is , which can be typed in as follows: (q*r)/(2*a)
Untitled picture.png Machine generated alternative text: A correct answer — A correct answer — , which can be typed in as follows: , which can be typed in as follows: 4mL • o ( -Q/ ) a epsi lon_e) A correct answer is , which can be typed in as follows: (q*r)/(2*a)

 

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