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Two particles \mathrm{A}$ and $\mathrm{B} having charges 20 \mu \mathrm{C}$ and $-5 \mu \mathrm{C} respectively are held fixed with a separation of 5 \mathrm{~cm}. At what position a third charged particle should be placed so that it does not experience a net electric force?

Option: 1 At 5 \mathrm{~cm}$ from $-5 \mu \mathrm{C} on the right side
 
Option: 2 At 5 \mathrm{~cm}$ from $20 \mu \mathrm{C} on the left side of system
 
Option: 3 At 1.25 \mathrm{~cm} from \mathrm{a}-5 \mu \mathrm{C} between two charges
 
Option: 4 At midpoint between two charges

Answers (1)

best_answer


Let c be the position where the net.force is zero (closer to the smaller charge for two unlike changes arrangement)
F_{A}= F_{B}
q\left ( E_{A} \right )= q\left ( E_{B} \right )
\frac{k\left ( 20\times 10^{-6} \right )}{\left ( 5+x \right )^{2}}= \frac{k\left ( 5\times 10^{-6} \right )}{x^{2}}
\frac{4}{\left ( 5+x \right )^{2}}= \frac{1}{x^{2}}
\frac{2}{5+x}= \frac{1}{x};2x= 5+x
  x= 5\, cm

Posted by

vishal kumar

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