Two point charges, \( q_1 = 36\,\mu\mathrm{C} \) and \( q_2 = -9\,\mu\mathrm{C} \) are placed at a distance of 30 cm. The distance from \( q_1 \), where the net electric field is zero, will be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 60 cm
For unlike charges the fields add between the charges, so the field can be zero only outside them, on the side of the smaller charge. Let the point be at distance \( x \) from \( q_1 \), beyond \( q_2 \).
\( \displaystyle \frac{k\,(36)}{x^{2}} = \frac{k\,(9)}{(x-30)^{2}} \;\Rightarrow\; \frac{x}{x-30} = \sqrt{\frac{36}{9}} = 2 \)
\( \displaystyle x = 2x - 60 \;\Rightarrow\; x = 60\,\mathrm{cm} \)
Shortcut: For two charges a distance \( d \) apart with a null point outside them, \( x/(x-d) = \sqrt{|q_1|/|q_2|} \), where \( x \) is the distance from the larger charge.