Maximum charge and voltage on a sphere

AI Thread Summary
The discussion centers on calculating the maximum charge and potential of a spherical conductor with a radius of 2 cm, where electrical breakdown occurs at an electric field of 3.0 x 10^6 Vm^-1. The maximum charge calculated using the electric field equation is 13μC, leading to a potential of 585kV on the sphere's surface. However, a correction reveals the radius should be 20 cm instead of 2 cm, which affects the calculations. The participants express uncertainty about the viability of the initial answers and the need to consider the correct radius. Accurate calculations are essential for determining the maximum charge and potential accurately.
shyguy79
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Hi guys, doing an assignment question

Homework Statement


A spherical conductor filled with air with a radius of 2cm is supported in the air. Electrical breakdown occurs when the electric field reaches 3.0 x 10^6 Vm^-1

(a) What is the maximum charge that can be placed on the sphere?
(b) What is the value of the potential on the surface of the conductor when carrying maximum charge

Homework Equations


E(r) = 1/4∏ε0 x Q/R^2 where 1/4∏ε0 = 9x10^9, Q = charge and R^2 =radius of sphere 0.2m
V(r) = 1/4∏ε0 x Q/R

The Attempt at a Solution


Ok, so for the (a) then using the equation for electric field E(r) rearranged for Q

Q = (E(r) x R^2)/9x10^9 =(3.0 x 10^6 Vm^-1 x 0.2^2)/9x10^9 = 13μC

Seems realistic? Just not sure of the next bit...

V(r) = 1/4∏ε0 x Q/R = 9x10^9 x (13x10-6C / 0.2m) = 585kV

585kV? Something has gone wrong... is this a viable answer or should I be taking into account the area of the sphere...
 
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shyguy79 said:
Hi guys, doing an assignment question

Homework Statement


A spherical conductor filled with air with a radius of 2cm is supported in the air. Electrical breakdown occurs when the electric field reaches 3.0 x 10^6 Vm^-1

(a) What is the maximum charge that can be placed on the sphere?
(b) What is the value of the potential on the surface of the conductor when carrying maximum charge

Homework Equations


E(r) = 1/4∏ε0 x Q/R^2 where 1/4∏ε0 = 9x10^9, Q = charge and R^2 =radius of sphere 0.2m
V(r) = 1/4∏ε0 x Q/R

The Attempt at a Solution


Ok, so for the (a) then using the equation for electric field E(r) rearranged for Q

Q = (E(r) x R^2)/9x10^9 =(3.0 x 10^6 Vm^-1 x 0.2^2)/9x10^9 = 13μC

Seems realistic? Just not sure of the next bit...

V(r) = 1/4∏ε0 x Q/R = 9x10^9 x (13x10-6C / 0.2m) = 585kV

585kV? Something has gone wrong... is this a viable answer or should I be taking into account the area of the sphere...
2cm = 0.02m , not 0.2m
 
Thanks for the heads up - that should have read

spherical conductor filled with air with a radius of 20cm
 

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