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Two positive charges Q and 4Q are placed...

Two positive charges Q and 4Q are placed at points A and B respectively, where B is at a distance d units to the right of A. The total electric potential due to these charges is minimum at P on the through A and B. What is (are) the distance(s) of P from A?

A

`(d)/(3)` units to the right of A

B

`(d)/(3)` units to the left of A

C

`(d)/(5)` units to the right of A

D

d uints to the left of A.

Text Solution

Verified by Experts

The correct Answer is:
a
Electric potential is minimum where `vecE_("total")=0`
At equilibrium `(KQ)/(x^(2))=(K4Q)/((d-x)^(2))rArr(1)/(x)=(2)/(d-x)rArrx=(d)/(3)`
`therefore` Total electric field will be zero at `(d)/(3)` units , right of A.
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