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I. Multiple Choice Questions · Q1

Q.Two identical point charges of magnitude −q-q are fixed as shown in the figure below. A third charge +q+q is placed midway between the two charges, at the point P. Suppose this charge +q+q is displaced a small distance from the point P, in the directions indicated by the arrows A1A_1, A2A_2 (along the line joining the two −q-q charges) and B1B_1, B2B_2 (perpendicular to that line). In which direction(s) will +q+q be stable with respect to the displacement?

(a) A1A_1 and A2A_2
(b) B1B_1 and B2B_2
(c) both directions
(d) No stable direction
two negative charges with a positive charge at the midpoint P, showing displacement directions A1, A2 along the line and B1, B2 perpendicular to it — Class 12 Physics electrostatics question
Figure
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✓ Free question

Step 1. At the midpoint P, the fields of the two equal −q-q charges cancel exactly, so +q+q sits in equilibrium there.

Step 2. Displace +q+q slightly along the line joining the two −q-q charges (directions A1A_1, A2A_2). Moving toward one −q-q brings +q+q closer to that charge and farther from the other; since +q+q is attracted toward each −q-q, the nearer charge now pulls harder than before and the farther one pulls less, so the net force pulls +q+q back toward P. This is a restoring force, so the equilibrium is stable along A1A_1, A2A_2.

Step 3. Displace +q+q slightly perpendicular to the line (directions B1B_1, B2B_2). By symmetry the two attractive forces from the −q-q charges now both acquire a component pushing +q+q further away from the line, since each −q-q pulls +q+q toward itself and the geometry tips both pulls in the same sideways sense away from P. This is a force that grows the displacement rather than opposing it, so the equilibrium is unstable along B1B_1, B2B_2.

Step 4. A configuration can be stable along one axis and unstable along the perpendicular axis simultaneously (a saddle point), which is exactly the case here.

✓Final answer

(a) A1A_1 and A2A_2

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