A positively charged ball hangs from a silk thread. If 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 is

Question

A positively charged ball hangs from a silk thread. If 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 is (A)  > F / q0 (B)  = F / q0 (C)  < F / q0 (D) cannot be estimated

Tardigrade Admin · Accepted Answer

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 / q0

Q. A positively charged ball hangs from a silk thread. If we put a positive test charge $q_{0}$ at a point and measure $F / q_{0}$ , then it can be predicted that the electric field strength $E$ is

$> F / q_{0}$

$= F / q_{0}$

$< F / q_{0}$

cannot be estimated

Q. A positively charged ball hangs from a silk thread. If 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 is

>F/q0​

=F/q0​

<F/q0​

cannot be estimated

Q. A positively charged ball hangs from a silk thread. If we put a positive test charge $q_{0}$ at a point and measure $F / q_{0}$ , then it can be predicted that the electric field strength $E$ is

$> F / q_{0}$

$= F / q_{0}$

$< F / q_{0}$