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 ______. (a) mg - qE (b) √(m²g² + q²E²) (c) mg + qE (d) √(m²g² - q²E²)

Answer: (b). At equilibrium the tension balances the resultant of mg (down) and qE (horizontal). These are perpendicular, so T = √((mg)² + (qE)²). So the answer is (b) √(m²g² + q²E²).

A simple pendulum has a bob with mass m and charge q. The pendulum… | JEE Main 2026 (22 Jan, Shift 1)

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Motion of Charges in an Electric Field

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 ______.

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Answer: (b)

At equilibrium the tension balances the resultant of $mg$ (down) and $qE$ (horizontal).
These are perpendicular, so $T = \sqrt{(mg)^2 + (qE)^2}$ .
So the answer is (b) $\sqrt{m^2g^2 + q^2E^2}$ .

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