A) \[>F/{{q}_{0}}\]
B) \[=F/{{q}_{0}}\]
C) \[<F/{{q}_{0}}\]
D) Cannot be estimated
Correct Answer: A
Solution :
Because of the presence of positive test charge q0 in front of positively charged ball, charge on the ball will be redistributed, less charge on the front half surface and more charge on the back half surface. As a result of this net force F between ball and point charge will decrease i.e. actual electric field will be greater than \[F/{{q}_{0}}.\]
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