Question

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Correct option is

$V=4πε_{0}1 [x6×10_{−6} −15−x(−4×10_{−6}) ]$

$0=4πε_{0}1 [x6×10_{−6} −15−x4×10_{−6} ]$

$0=x6×10_{−6} −15−x4×10_{−6} $

$x6×10_{−6} =15−x4×10_{−6} $

$⇒6(15−x)=4x$

$⇒90−6x=4x$

$⇒10x=90$

$⇒x=1090 =9cm$

The other possibility is that point of zero potential $P$ may lie on $AB$ produced at a distance $x$ from the charge $+6×10_{−6}C$ at $A$ as shown in the figure.

Potential at $P$

$V=4πε_{0}1 [x6×10_{−6} +(15−x)(−4×10_{−6}) ]$

$0=4πε_{0}1 [x6×10_{−6} −(x−15)4×10_{−6} ]$

$0=x6×10_{−6} −(x−15)4×10_{−6} $

$⇒x6×10_{−6} =x−154×10_{−6} $

$⇒x6 =x−154 $

$⇒6x−90=4x$

$⇒2x=90$

$⇒x=290 =45cm$

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