Maximum charge on a spherical capacitor

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
  • #1
lorenz0
148
28
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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  • #2
Why are you using the potential at all? The limit is given in terms of the E field.
Where is it biggest?
 
  • #3
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?
 
  • #4
Yes.
 
  • #5
t
hutchphd said:
Yes.
Thank you.
 
  • Like
Likes hutchphd

1. What is the maximum charge that can be stored on a spherical capacitor?

The maximum charge that can be stored on a spherical capacitor is dependent on the size and material of the capacitor. However, in theory, the maximum charge is infinite as long as the electric field strength does not exceed the breakdown strength of the material.

2. How is the maximum charge on a spherical capacitor calculated?

The maximum charge on a spherical capacitor can be calculated using the formula Q = 4πε0R2V, where Q is the charge, ε0 is the permittivity of free space, R is the radius of the capacitor, and V is the potential difference between the two plates.

3. Can the maximum charge on a spherical capacitor be exceeded?

In theory, the maximum charge on a spherical capacitor can be exceeded if the electric field strength exceeds the breakdown strength of the material. However, this can cause damage to the capacitor and is not recommended.

4. How does the distance between the plates affect the maximum charge on a spherical capacitor?

The distance between the plates does not directly affect the maximum charge on a spherical capacitor. However, a larger distance between the plates can result in a weaker electric field, which can affect the maximum charge that can be stored.

5. What factors can affect the maximum charge on a spherical capacitor?

The maximum charge on a spherical capacitor can be affected by the size and material of the capacitor, the potential difference between the plates, and the surrounding environment. Temperature and humidity can also have an impact on the maximum charge that can be stored on a capacitor.

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