Why is electric field strength greater than F/q₀ for a charged ball?

[sainiroop]

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...

 

[TxRationalist]

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.

 

[collinsmark]

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).