Question

Two particles A and B having charges $+2\times {10}^{\circ }C$ and $-4\times {10}^{-6}C$ respectively are held fixed at a separation of 20 cm The point where electric potential becomes zero is

• A. $\frac{20}{3}$ cm
• B. $\frac{20}{5}$ cm
• C. $\frac{20}{7}$ cm
• D. $\frac{20}{9}$ cm

Answer 1

The correct option is A $\frac{20}{3}$ cm

$\begin{array}{c}Sincetwocheareofopposite\sin g,sothepointatwhichelectricfieldiszeroliesoutsidethesystem\\ Letxcmbethedistanceofthepointfrom2\times {10}^{-6}C.\\ therefore,\\ \frac{k\times 2\times {10}^{-6}}{x^2}=\frac{k\times 4\times {10}^{-6}}{{\left(20+x\right)}^2}\\ Weget\\ x=20\left(1\pm \sqrt{2}\right)\\ Since,pointsliesoutsidethesystem,wetakeonlypositivesign.\\ x=20\left(1+\sqrt{2}\right)cm\\ letdbethedis\tan cefromtheche2\times {10}^{-6}Catwhichpotentialiszero.\sin cepotentialisscalerquantiy.\\ Thispointsliesinsidethesystem.\\ \frac{k\times 2\times {10}^{-6}}{d}+\frac{k\times \left(-4\times {10}^{-6}\right)}{20-d}=0\\ weget\\ d=\frac{20}{3}cm\\ Hence,optionAisthecorrectanswer.\end{array}$