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Standard XII

Physics

Electrostatic Shielding

A positively ...

Question

A positively charged ball hangs from a silk thread. 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.

> F / q_{0}

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= F / q

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< F / q_{0}

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Cannot be estimated

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Solution

The correct option is B $> F / q_{0}$
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 change 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 $F / q_{0}$ .

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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.

A positively charged ball hangs from a silk thread. 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.