Capacitance and max field strength of a concentric sphere capacitor

AI Thread Summary
The discussion focuses on calculating the capacitance and maximum electric field strength of a concentric sphere capacitor with given dimensions and a liquid insulator. The capacitance is derived using the formula adjusted for relative permittivity, resulting in a value of 133.4 pF. To find the maximum electric field strength, the charge is calculated from the capacitance and potential difference, leading to an electric field strength of 2.14. Clarifications confirm that the maximum electric field just outside the inner sphere is equivalent to the electric field in the gap between the spheres. The calculations and concepts discussed align with standard physics principles for capacitors.
Andzus
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Homework Statement


A capacitor is formed by two concentric spheres r1= 30 cm and r2=15 cm, with a liquid insulant between them with relative permittivity Er=4. The potential difference between the spheres is 1 kV.
Find:
1. capacitance
2. maximum field strength in the gap between the two spheres

Homework Equations



C=4*pi*epsilon/[1/r1-1/r2]
Q=C*potential_difference
Emax=Q/4*pi*epsilon*r^2


The Attempt at a Solution


1. To get the capacitance I added Er=4 to the formulae
C=4*pi*Er*epsilon/[1/r1-1/r2]
and got C=133.4 pF, is this right? I added Er=4 because we usually use Er=1[for air]

2. to get the max field strength I substited Q=C*potential_difference into Emax and used r=r1-r2. So now the equation is:
Emax=C*potential_difference/4*pi*Er*epsilon*(r1-r2)^2
and got Emax=2.14
 
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You have a value for capacitance, so you should be able to find the charge on the capacitor given the voltage difference.

Given the charge, the electric field just outside the 15cm radius sphere can be found via Gauss' law.

E = \frac{Q}{4 \pi \epsilon_0 \epsilon_r r^2}
 
gneill said:
You have a value for capacitance, so you should be able to find the charge on the capacitor given the voltage difference.

Given the charge, the electric field just outside the 15cm radius sphere can be found via Gauss' law.

E = \frac{Q}{4 \pi \epsilon_0 \epsilon_r r^2}

Just to clarify - is the maximum electric field just outside the 15cm radius sphere the same as the electric field in the gap between the two spheres?
 
Andzus said:
Just to clarify - is the maximum electric field just outside the 15cm radius sphere the same as the electric field in the gap between the two spheres?

Yes it is. The variable of interest in the equation I provided is r, and it ranges from just above the surface of the inner sphere to just before the inner surface of the outer sphere.
 
gneill said:
Yes it is. The variable of interest in the equation I provided is r, and it ranges from just above the surface of the inner sphere to just before the inner surface of the outer sphere.

Thank You very much gneill! :)
 
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