Q66 / 75 · JEE Main 2026 (22 January Shift 1) · Physics
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)
- Correct answerA$\sqrt{m^{2} g^{2}+q^{2} E^{2}}$
- B$m \mathrm{~g}+q E$
- C$m g-q E$
- D$\sqrt{m^{2} g^{2}-q^{2} E^{2}}$
Solution
The charged bob experiences two forces at equilibrium:
Gravitational force: $F_g = mg$ (downward)
Electric force: $F_E = qE$ (horizontal, in direction of field)
These forces are perpendicular to each other. The string tension must balance their resultant.
At equilibrium, the string makes angle $\theta$ with vertical such that:
Vertical: $T\cos\theta = mg$
Horizontal: $T\sin\theta = qE$
The net force from gravity and electric field is $\sqrt{(mg)^2 + (qE)^2}$
Therefore, the tension in the string is: $T = \sqrt{m^2g^2 + q^2E^2}$
Gravitational force: $F_g = mg$ (downward)
Electric force: $F_E = qE$ (horizontal, in direction of field)
These forces are perpendicular to each other. The string tension must balance their resultant.
At equilibrium, the string makes angle $\theta$ with vertical such that:
Vertical: $T\cos\theta = mg$
Horizontal: $T\sin\theta = qE$
The net force from gravity and electric field is $\sqrt{(mg)^2 + (qE)^2}$
Therefore, the tension in the string is: $T = \sqrt{m^2g^2 + q^2E^2}$
Hence, the correct answer is option: 'A'.