Maximum charge and voltage on a sphere

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SUMMARY

The maximum charge that can be placed on a spherical conductor with a radius of 20 cm, before electrical breakdown occurs at an electric field of 3.0 x 10^6 Vm^-1, is calculated to be 13 μC. The potential on the surface of the conductor when carrying this maximum charge is determined to be 585 kV. The calculations utilize the equations E(r) = 1/4∏ε0 x Q/R^2 and V(r) = 1/4∏ε0 x Q/R, with the constant 1/4∏ε0 being 9 x 10^9. It is crucial to use the correct radius of 0.2 m for accurate results.

PREREQUISITES
  • Understanding of electric fields and potentials
  • Familiarity with the equations for electric field and potential of a sphere
  • Knowledge of the constants involved in electrostatics, specifically 1/4∏ε0
  • Ability to perform unit conversions, particularly between centimeters and meters
NEXT STEPS
  • Study the derivation and applications of the equations for electric fields and potentials in electrostatics
  • Learn about the implications of electrical breakdown in different materials
  • Explore the concept of capacitance in spherical conductors
  • Investigate the effects of varying the radius of a conductor on charge and potential
USEFUL FOR

Students studying electrostatics, physics educators, and anyone involved in electrical engineering or related fields seeking to understand the behavior of charged spherical conductors.

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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