Finding a relation

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Homework Statement


The capacitance of a spherical condenser is 1uF. If the spacing between the two spheres is 1mm, then what is the radius of the outer sphere

Homework Equations


[/B]
C = 4 pi Epsilon * ab/b-a (for a spherical capacitor)



The Attempt at a Solution



Given C = 1uF (u = 10^-6)
b-a = 10^-3

1uF = 4 pi Epsilon * ab/10^-3
On solving
9 = ab

How do I proceed forward from here? Please help
 

Answers and Replies

  • #2
haruspex
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Homework Statement


The capacitance of a spherical condenser is 1uF. If the spacing between the two spheres is 1mm, then what is the radius of the outer sphere

Homework Equations


[/B]
C = 4 pi Epsilon * ab/b-a (for a spherical capacitor)



The Attempt at a Solution



Given C = 1uF (u = 10^-6)
b-a = 10^-3

1uF = 4 pi Epsilon * ab/10^-3
On solving
9 = ab

How do I proceed forward from here? Please help
You have two equations and two unknowns.
Apply the standard method of solving simultaneous equations. Use one equation to express one variable in terms of the other, then substiute for it in the other equation.
 
  • #3
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You have two equations and two unknowns.
Apply the standard method of solving simultaneous equations. Use one equation to express one variable in terms of the other, then substitute for it in the other equation.
Yes I did but this is what happened;
ab = 9 -------- (1)
b-a = 10^-3
b = a + 10^-3 -- substituting in (1)

a(a+10^-3) = 9
Still stuck! :(
 
  • #4
haruspex
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Yes I did
Next time, please post your working as far as you get in the first post.
a(a+10^-3) = 9
Still stuck! :(
It's just a quadratic equation. But it should be obvious what a will be approximately.
By the way, I did not check your derivation of ab=9.
 
  • #5
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Next time, please post your working as far as you get in the first post.

It's just a quadratic equation. But it should be obvious what a will be approximately.
By the way, I did not check your derivation of ab=9.
I never saw that quadratic equation coming! Thank you. This it!
 

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