Max Charge Capacity on a Spherical Body: Calculating the Limit

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SUMMARY

The maximum charge that can be stored on a spherical body of radius r is determined using the formula Q = 4 π ε r^2 E(r), where E(r) represents the maximum electric field strength. The capacitance of the sphere is given by C = 4 π ε r, and the relationship between charge (Q), capacitance (C), and voltage (V) is expressed as Q = CV. The breakdown voltage of the surrounding medium, such as air, must be considered to ensure that the voltage does not exceed E0, the breakdown threshold.

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Me again.

Is there a formula for the maximum amount of charge may be stored on a body (assume spherical) of radius r? I assume radius is the only factor, though mass is known also. I just don't remember seeing anything for this and can't find it in all my books for the life of me.

Thanks.
 
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Lets try ::
Capacitance:: C=[tex]4 \pi \epsilon r[/tex]
Q=CV
[tex]Q=4 \pi \epsilon r V[/tex]
Now V should be less than E0 Breakdown Voltage
so max can be
[tex]Q_{max} = 4 \pi \epsilon r E_0[/tex]
Well I am not too sure abt above but its a try , if u can provide some other logic it will be welcomed
 
himanshu121 said:
Lets try ::
Capacitance:: C=[tex]4 \pi \epsilon r[/tex]
Q=CV
[tex]Q=4 \pi \epsilon r V[/tex]
Now V should be less than E0 Breakdown Voltage
so max can be
[tex]Q_{max} = 4 \pi \epsilon r E_0[/tex]
Well I am not too sure abt above but its a try , if u can provide some other logic it will be welcomed
Dammit, sorry I was being dumb. I didn't notice we were given the electrical breakdown of air in the problem, so I know the maxmum electric field, I know the radius, and from this I can easily get the charge:

Q = [tex]4 \pi \epsilon r^2[/tex]E(r)

Apologies for wasting thine time, and thanks for posting back. I was pretty tired. :zzz: Take care.
 

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