A positively charged ball hangs from a silk thread. We put a positive test charge q₀ at a point and measure F / q₀ , then it can be predicted that the electric field strength E

Step-by-step solution

Due to presence of test charge q₀ 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 / q₀ .