Maximum charge and voltage on a sphere

In summary: Electrical breakdown occurs when the electric field reaches 3.0 x 10^6 Vm^-1In summary, the sphere can hold a maximum of 13μC of charge.
  • #1
shyguy79
102
0
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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  • #2
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
 
  • #3
Thanks for the heads up - that should have read

spherical conductor filled with air with a radius of 20cm
 

1. What is the maximum charge that can be placed on a sphere?

The maximum charge that can be placed on a sphere depends on the material and size of the sphere. For conductive spheres, the maximum charge is limited by the breakdown voltage of the material, while for non-conductive spheres, the maximum charge is limited by the dielectric strength of the material.

2. How is the maximum charge on a sphere calculated?

The maximum charge on a sphere can be calculated using the formula Q = 4πε₀RΔV, where Q is the maximum charge, ε₀ is the permittivity of free space, R is the radius of the sphere, and ΔV is the breakdown voltage or dielectric strength of the material.

3. What is the relationship between maximum charge and voltage on a sphere?

The maximum charge and voltage on a sphere are directly proportional. This means that as the voltage increases, the maximum charge that can be placed on the sphere also increases. However, there is a limit to how much charge can be placed on a sphere, even with increasing voltage.

4. Can the maximum charge and voltage on a sphere be exceeded?

Yes, it is possible to exceed the maximum charge and voltage on a sphere, but this can result in the breakdown of the material and potentially cause damage. It is important to adhere to the maximum charge and voltage limits to ensure safety and proper functioning of the sphere.

5. How does the shape of a sphere affect its maximum charge and voltage?

The shape of a sphere does not significantly affect its maximum charge and voltage. As long as the material and size remain constant, the maximum charge and voltage will not be greatly affected by the shape of the sphere.

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