Find Radius of Outer Sphere for 1uF Spherical Condenser

[James2911]

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

 

[haruspex]

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.

 

[James2911]

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! :(

 

[haruspex]

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.

 

[James2911]

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!