Two point charges of \( 1 \times 10^{-6}\,\mathrm{C} \) and \( 4 \times 10^{-6}\,\mathrm{C} \) are placed at a certain distance away from each other. The resultant electric field intensity is zero at a distance of 15 cm from the charge \( 1 \times 10^{-6}\,\mathrm{C} \). The distance between the two point charges will be:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 45\,\mathrm{cm} \)
Both charges are positive, so their fields oppose each other only at a point between them. Let the point be at \( x = 15\,\mathrm{cm} \) from \( q_1 \) and let the charges be \( d \) apart. The fields cancel there:
\( \displaystyle \frac{k q_1}{x^{2}} = \frac{k q_2}{(d - x)^{2}} \)
\( \displaystyle \frac{(d - x)^{2}}{x^{2}} = \frac{q_2}{q_1} = 4 \;\Rightarrow\; \frac{d - x}{x} = 2 \)
\( \displaystyle d - x = 2 \times 15\,\mathrm{cm} = 30\,\mathrm{cm} \)
\( \displaystyle d = 15\,\mathrm{cm} + 30\,\mathrm{cm} = 45\,\mathrm{cm} \)