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.
lorenz0
Messages
151
Reaction score
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
 
Physics news on Phys.org
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top