Question 39 of 42
Q.Two point charges are placed 0·18 m apart in air. One charge is four times the other charge. If the electric field is zero at a point on the line joining the two charges, then the position of the point is
(a) on the extended part of the line of the charges and at a distance of 0·06 m from the larger charge.
(b) between the two point charges on the line and at a distance of 0·06 m from the smaller charge.
(c) between the two point charges on the line and at a distance of 0·04 m from the larger charge.
(d) on the extended part of the line and at a distance of 0·04 m from the smaller charge.
West Bengal WbchseWest Bengal HS (WBCHSE) Board 2026MCQ· 1mImportance★★★★★
93% · 39/42 Questions
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Start your 14-day free trial to unlock the full solution →For like charges the null point lies between them, nearer the smaller charge. Solving kq/x² = 4kq/(0·18−x)² gives x = 0·06 m from the smaller charge. Option (b).
Let the smaller charge be q and the larger 4q, separated by d = 0·18 m. The neutral point (E = 0) lies between them (both being of the same sign) at distance x from q.
Step 1 — equate field magnitudes:
kq/x² = k(4q)/(d−x)².
Step 2 — cancel kq and take square roots:
(d−x)²/x² = 4 → (d−x)/x = 2 → d − x = 2x → d = 3x.
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