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Question

# Two point charges (+Q) and (−2Q) are fixed on the X-axis at positions a and 2a from origin respectively. At what positions on the axis, the resultant electric field is zero.

A
Only x=2a
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B
Only x=2a
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C
Both x=±2a
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D
x=3a2 only
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Solution

## The correct option is B Only x=−√2aBy intuition we know that it will be at two point i.e., A and B.Let distance of A from (a,0) be x and distance of B from (2a,0) be y.Point A:k(+Q)x2+k(−2Q)(a+x)2=01x2−2(a+x)2=0(a+x)2=2x2a2+x2+2ax=2x2x2−2ax−a2=0x=2a±√4a2+4a22=2a±√8a22=2a±2√2a2x=(1±√2)a⇒x=(1+√2)aPoint B:K(+Q)(a+y)2+K(−2Q)y2=01(a+y)2=2y2y2=2(a+y)2y2=2a2+2y2+4ayy2+4ay+2a2=0y=−4a±√16a2−4a22=−4a±√8a22=−4a±2√2a2y=(−2±√2)a⇒y= Not feasible as both are negative.Thus, as x is found, accordingly the point on x-axis will be (−√2a,0).

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