Two particles A and B having charges \( 8\times10^{-6}\,\mathrm{C} \) and \( -2\times10^{-6}\,\mathrm{C} \) respectively are held fixed with a separation of 20 cm. At what distance from the charge \( -2\times10^{-6}\,\mathrm{C} \) should a third charged particle be placed so that it does not experience any net electric force?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 20 cm
The third charge feels no net force where the Coulomb forces of A and B on it are equal and opposite.
A and B carry opposite charges, so between them both forces point the same way.
So the null point lies outside the pair, on the side of the smaller charge B. There, A is farther away, and this balances its larger charge.
Let x be the distance of the third charge q beyond B, so its distance from A is \( 0.20\,\mathrm{m}+x \). Equating the magnitudes of the two forces:
\( \displaystyle \frac{k\,(8\times10^{-6}\,\mathrm{C})\,q}{(0.20\,\mathrm{m}+x)^2}=\frac{k\,(2\times10^{-6}\,\mathrm{C})\,q}{x^2} \)
\( \displaystyle \left(\frac{0.20\,\mathrm{m}+x}{x}\right)^2=4 \)
\( \Rightarrow\; 0.20\,\mathrm{m}+x=2x \)
\( x=0.20\,\mathrm{m}=20\,\mathrm{cm} \)