A simple pendulum has a bob with mass $m$ and charge q . The pendulum string has negligible mass. When a uniform and horizontal electric field $\vec{E}$ is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is \_\_\_\_ . (g : acceleration due to gravity)

(C) $\sqrt{m^{2} g^{2}+q^{2} E^{2}}$

(D) $\sqrt{m^{2} g^{2}-q^{2} E^{2}}$

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Laws of Motion

A simple pendulum has a bob with mass

m

and charge q . The pendulum string has negligible mass. When a uniform and horizontal electric field

E

is applied, the tension in the string changes. The final tension in the string, when pendulum attains an equilibrium position is ____ .
(g : acceleration due to gravity)

(A) $m g - q E$
(B) $m g + q E$
(C) $m^{2} g^{2} + q^{2} E^{2}$ ✓ Correct
(D) $m^{2} g^{2} - q^{2} E^{2}$

Answer: (C)

Solution

At equilibrium the bob is held by three forces: its weight

m g

acting downward, the horizontal electric force

q E

, and the string tension

T

. The string aligns itself along the resultant of the two applied forces, which are mutually perpendicular, so the tension equals their vector sum:

T = (m g)^{2} + (q E)^{2} .

JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Single correct option

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