Electric Potential of a Sphere at Different Locations

  • #1
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Homework Statement


A solid insulating sphere of radius a = 3.6 cm is fixed at the origin of a co-ordinate system as shown. The sphere is uniformly charged with a charge density ρ = -215 μC/m3. Concentric with the sphere is an uncharged spherical conducting shell of inner radius b = 11 cm, and outer radius c = 13 cm.

I computed the right values for Question 2) and 3), but for Question 2
I don't understand why I have to integrate over infinity to c = 13 cm, why can't I integrate from the origin of the insulating sphere to b since Question 2 is asking for inner surface of conducting shell?

Question 3)
I don't understand why I have to integrate E potential from infinity to C then add integration from B to A.
Couldn't I just integrate from the origin to a to get the outer surface? Is it because the potential is zero at infinity which is the starting point?

Homework Equations


Integral of (electric field dl)

The Attempt at a Solution


Question 2) k*(-215*10^-6)(4/3pi(0.0036)^3) * 1/c
Question 3) k*(-215*10^-6)(4/3pi(0.0036)^3) * 1/c + k*(-215*10^-6)(4/3pi(0.0036)^3) * (1/a - 1/b)


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Answers and Replies

  • #2
haruspex
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why can't I integrate from the origin of the insulating sphere to b since Question 2 is asking for inner surface of conducting shell?
Electric potential is always relative to some defined zero. In principle, you can choose that zero arbitrarily, but in most cases the convention is to define it as the potential at infinity. In the present case you are instructed to use that convention.
This means that the potential at, say, the origin need not be zero. Integrating from the origin to the inner surface of the shell will only tell you the relative potential of those two points. To find the potential relative to the defined zero you must do at least one integral to infinity.
 

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