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Due to an electric charge Q , field in...

Due to an electric charge Q , field intensity at the position of test charge `q_0` is ` vecE`. If the test charge is replaced by `-q_0`, then field intensity becomes

A

`-q_(0)vec E`

B

`(vecE)/(-q_(0))`

C

`0`

D

`vecE`

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CHHAYA PUBLICATION-ELECTRIC FIELD-EXERCISE ( MULTIPLE CHOICE QUESTION )
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