Maximum charge on a spherical capacitor

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SUMMARY

The maximum charge on a spherical capacitor is determined by the relationship between the electric field and the charge on the inner sphere. The electric field generated by the charge +Q is expressed as ##\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}##, and when a dielectric is present, it modifies to ##\vec{E}_d=\frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{Q}{r^2}\hat{r}##. The maximum charge, ##Q_{max}##, can be calculated using the formula ##Q_{max}=E_{max}4\pi\varepsilon_0\varepsilon_rR_1^2##, where ##E_{max}## is the maximum electric field at the surface of the inner sphere, ##R_1##. This establishes a clear relationship between charge, electric field, and the geometry of the capacitor.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with spherical capacitors
  • Knowledge of dielectric materials and their properties
  • Basic calculus for evaluating integrals
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  • Study the derivation of electric fields in capacitors
  • Learn about the effects of different dielectric materials on capacitance
  • Explore the mathematical implications of Gauss's Law in spherical symmetry
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lorenz0
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Homework Statement
A spherical capacitor has internal radius ##R_1## and external radius ##R_2##.
Between the spheres there is a dielectric with constant ##\varepsilon_r##.
If the maximum electric field that can be applied without electrical discharges occurring is ##E_{max}##, find the corresponding maximum charge that can be put on the plates.
Relevant Equations
##\Delta V=\int \vec{E}\cdot d\vec{l}##
The electric field is the one generated by the charge ##+Q## on the inner sphere of the capacitor, which generates a radial electric field ##\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}## which, due to the presence of the dielectric, become ##\vec{E}_d=\frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{Q}{r^2}\hat{r}## so ##\Delta V=\int \vec{E}_d\cdot d\vec{l}=\int_{R_1}^{R_2}\frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{Q}{r^2}\hat{r}\cdot d\vec{l}=\frac{Q}{4\pi\varepsilon_0\varepsilon_r}\frac{R_2-R_1}{R_1R_2}.##

So, ##E=\frac{\Delta V}{R_2-R_1}=\frac{Q}{4\pi\varepsilon_0\varepsilon_r}\frac{1}{R_1R_2}\leq E_{max}\Rightarrow \frac{Q_{max}}{4\pi\varepsilon_0\varepsilon_r}\frac{1}{R_1R_2}= E_{max}\Leftrightarrow Q_{max}=E_{max}4\pi\varepsilon_0\varepsilon_rR_1R_2##.

Does this make sense? Thanks
 
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Why are you using the potential at all? The limit is given in terms of the E field.
Where is it biggest?
 
hutchphd said:
Why are you using the potential at all? The limit is given in terms of the E field.
Where is it biggest?
##E## is biggest on the surface of the inner sphere so ##E_{max}=E(R_1)=\frac{Q_{max}}{4\pi\varepsilon_0\varepsilon_r R_1^2}\Leftrightarrow Q_{max}=4\pi\varepsilon_0\varepsilon_r R_1^2 E_{max}##. Is this correct?
 
Yes.
 
t
hutchphd said:
Yes.
Thank you.
 
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