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Lera25 [3.4K]
3 years ago
13

What minimum radius must a large conducting sphere of an electrostatic generating machine have if it is to be at 20000 V without

discharge into the air?
How much charge will it carry?
Physics
1 answer:
zmey [24]3 years ago
6 0
First of all,

The electric field at the surface of the sphere is given by


 E = kQ/r²

The field strength at which breakdown occurs in the air is <span>3.0 MV/m
</span>
So,  E = 3.0 MV/m

<span>The sphere potential is defined as

 V = kQ/r</span>

<span>If we divide E/V  we get

 E = V/r </span>

<span>r = V/E  = 20000V / 3.10^6 V/m  =  6.66 </span>10^-3 m = 6.66 mm

2. Charge

<span>V = kQ/r .............>>
</span>
Q = Vr/k = 20000V *( 6.66 10^-3 m)/ (9.10^9 N m2/C2) = 1.481 10^-8 C

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