# Maximum charge on a spherical capacitor

• lorenz0
In summary, the electric field generated by the charge ##+Q## on the inner sphere of a capacitor is given by ##\vec{E}=\frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{r}##. When a dielectric is present, the field becomes ##\vec{E}_d=\frac{1}{4\pi\varepsilon_0\varepsilon_r}\frac{Q}{r^2}\hat{r}##. The maximum electric field, ##E_{max}##, is determined by the limit ##\frac{Q_{max}}{4\pi\varepsilon_0\varepsilon_r}\frac{1

#### lorenz0

Homework Statement
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

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.

hutchphd