A positively charged ball hangs from a silk thread. We put a...

Due to presence of test charge $q_0$ in front of positively charged ball, there would be a redistribution of charge on the ball. In the redistribution of charge, there will be less charge on front half surface and more charge on the back half surface. As a result, the net force $F$ between ball and charge will decrease, i.e. the electric field is decreased. Thus, actual electric field will be greater than $F / q_0$.