Physics Electrostatics Question

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SUMMARY

The discussion centers on the calculation of electric field strength (E) in relation to a positive test charge (q0) placed near a positively charged ball. The consensus is that the electric field strength (E) is greater than the force per unit charge (F/q0), confirming option (a) as the correct answer. Participants emphasize the importance of Coulomb's law and the definition of electric field in deriving this conclusion, particularly in the context of uniformly charged spheres and the inverse square law.

PREREQUISITES
  • Coulomb's Law
  • Understanding of Electric Field Concepts
  • Knowledge of Test Charge Definition
  • Familiarity with Inverse Square Law in Electrostatics
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  • Study Coulomb's Law and its applications in electrostatics
  • Explore the concept of electric field strength and its calculations
  • Investigate the properties of uniformly charged spheres
  • Learn about the implications of the inverse square law in electric fields
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sainiroop
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A positive charged ball hangs from a silk thread. We put a positive test charge q0 at a point and measure F/q0, It can be predicted that E (elect field strength) becomes :
a) greater than f/q0
b) equal to f/q0
c) less than F/q0
d) cannot be estimated.

Answer (Given) : a) greater than f/q0

Need an explanation for the above...
 
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I would like some enlightenment too. from what i remember, outside a uniformly charged sphere of radius R the Electric field falls off according the classical inverse square E-field law with the sphere acting as a point charge...but it's been a while for me.
 
sainiroop said:
Answer (Given) : a) greater than f/q0

Need an explanation for the above...

Hello sainiroop,

Are you saying that is the answer you gave, or is that the answer the instructor or textbook gave?

The way I would approach this question is to first write down Coulomb's law, and then to ask myself, what is the definition of Electric field? Putting the two together, the answer should become obvious (also considering the definition of test charge, as opposed to just any old charge). :wink:
 

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