Maximum charge on a spherical capacitor

AI Thread Summary
The discussion focuses on the relationship between charge, electric field, and potential in a spherical capacitor. The electric field generated by the charge on the inner sphere is expressed as E = (1/(4πε₀εᵣ))(Q/r²) and the potential difference is calculated using this field. The maximum electric field occurs at the surface of the inner sphere, leading to the conclusion that Q_max can be determined as Q_max = 4πε₀εᵣR₁²E_max. The conversation confirms the mathematical derivations and clarifies the significance of using potential in the context of electric fields. Overall, the discussion emphasizes the interplay between charge, electric field strength, and capacitor geometry.
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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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