A) \[-E\]
B) \[E/3\]
C) \[-3E\]
D) \[-E/3\]
Correct Answer: B
Solution :
The field produced by charge ? 3Q at A, this is E as mentioned in the Example. \[\therefore \]\[E=\frac{3Q}{{{x}^{2}}}\] (along AB directed towards negative charge) Now field at location of ? 3Q i.e. field at B due to charge Q will be \[E'=\frac{Q}{{{x}^{2}}}=\frac{E}{3}\] (along AB directed away from positive charge)You need to login to perform this action.
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