The figure shows the electric field lines in the vicinity of two poin…

The correct answer is Option (2) → The magnitude of the ratio \( q_1/q_2 \) is more than 1

Electric field lines start from positive charges and end at negative charges. By Gauss’s law, the flux through a closed surface around a charge is \( q/\varepsilon_0 \). The flux is proportional to the number of field lines. So a larger charge has more lines leaving or entering it.

Option (1) \( q_1 \) is negative and \( q_2 \) is positive — Incorrect. From the figure, the arrows point away from \( q_1 \) and into \( q_2 \), so \( q_1 \) is positive and \( q_2 \) is negative.

Option (2) The magnitude of the ratio \( q_1/q_2 \) is more than 1 — Correct. From the figure, many more lines leave \( q_1 \) than end on \( q_2 \). Several lines from \( q_1 \) go off without reaching \( q_2 \). Hence \( |q_1|\gt|q_2| \), that is, \( |q_1/q_2|\gt1 \).

Option (3) \( q_1 \) and \( q_2 \) have the same polarity — Incorrect. The lines run from \( q_1 \) to \( q_2 \). This happens only between charges of opposite sign. The lines of two like charges bend away from each other.

Option (4) The electric field is strongest midway between the charges — Incorrect. Midway, the lines are spread far apart, so the field there is weak. The lines crowd together close to each charge, and the field is strongest there.

NCERT (2026-27):
Electric Flux – “We now come to the definition of electric flux. Electric flux Δφ through an area element ΔS is defined by

    \( \Delta\phi=\mathbf{E}\cdot\Delta\mathbf{S}=E\,\Delta S\cos\theta \qquad (1.11) \)

which, as seen before, is proportional to the number of field lines cutting the area element.”

Gauss’s Law – “We state Gauss’s law without proof:

    \( \text{Electric flux through a closed surface }S=q/\varepsilon_0 \qquad (1.31) \)

q = total charge enclosed by S.”