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A pendulum with a charged bob between two charged plates looks like an electrostatics problem, but it is really a gravity problem in disguise. Once you see that a uniform electric field just adds a constant force, every version reduces to one formula. This post builds that formula, works through four set-ups with one set of numbers, and shows the trap where the field is not uniform.
Is this in the syllabus?
Yes, as a combination of two listed topics. The JEE (Main) 2026 syllabus and the NEET (UG) 2026 syllabus list "simple pendulum: derivation of expression for its time period" in Unit 10 and the electric field due to a point charge in Unit 11. Section 13.8 of the NCERT Class 11 Physics chapter Oscillations (Reprint 2026-27) derives T = 2π√(L/g) for small swings, and Eq. 1.8 of the NCERT Class 12 chapter Electric Charges and Fields gives the force on a charge as F = qE.
What is effective g for a charged bob?
In a uniform field, qE is the same at every point of the swing, just like mg. So the two add into one constant force:
m g_eff = mg + qE, so g_eff = g + qE/m (as vectors)
The bob hangs along g_eff, and NCERT's derivation goes through with g replaced by g_eff:
T = 2π√(L / g_eff)
If q is negative, qE points opposite to E. Write the force direction before anything else.
Worked example: one pendulum, four fields
A bob of mass 20 g carries +5.0 µC on a 1.0 m string. Take g = 9.8 m/s² and E = 2.0 × 10⁴ N/C.
The electric force is qE = 5.0 × 10⁻⁶ × 2.0 × 10⁴ = 0.10 N, so qE/m = 0.10 ÷ 0.020 = 5.0 m/s².
No field. T = 2π√(1.0/9.8) = 2.01 s.
Field downward. The force adds to gravity: g_eff = 9.8 + 5.0 = 14.8 m/s². T = 2π√(1.0/14.8) = 1.63 s, shorter.
Field upward. The force opposes gravity: g_eff = 9.8 − 5.0 = 4.8 m/s². T = 2π√(1.0/4.8) = 2.87 s, longer. A negative charge in a downward field gives the same answer.
Field horizontal. The two are at right angles: g_eff = √(9.8² + 5.0²) = 11.0 m/s². T = 2π√(1.0/11.0) = 1.89 s. The string tilts from the vertical by tan⁻¹(5.0/9.8) ≈ 27°, and the tension at rest is m g_eff = 0.020 × 11.0 = 0.22 N.
What field stops the pendulum swinging?
If qE exactly cancels mg, g_eff = 0 and there is no restoring force. For our bob, the field must point up with E = mg/q = (0.020 × 9.8) ÷ (5.0 × 10⁻⁶) = 3.9 × 10⁴ N/C. Displaced, the bob would stay where you put it. A stronger upward field makes g_eff point up, and the pendulum hangs, and swings, upside down above the support.
What if the charge is at the point of suspension?
A classic JEE trap: a second charge is fixed at the point where the string is tied. Its force on the bob is along the string at every instant, because the bob is always at distance L from it. A force along the string cannot pull the bob back towards the centre of its swing. It only changes the tension. So the period stays 2π√(L/g).
This is not a uniform field, so g_eff does not apply. Always check whether the field is uniform before you use the shortcut.
Which mistakes cost marks?
Ignoring the sign of q. A negative charge reverses the force.
Forgetting to divide by m. qE is a force; g_eff needs qE/m, an acceleration.
Adding horizontal fields as numbers. At right angles, use Pythagoras.
Using g_eff for a point charge at the pivot. The period is unchanged there.
Large swings. NCERT's result assumes small angles, where sin θ ≈ θ.
Unit slips. 20 g is 0.020 kg; 5.0 µC is 5.0 × 10⁻⁶ C.
The same idea, with acceleration in place of an electric force, appears in our post on the pendulum in a lift or accelerating car. For a charge moving freely through a field, see electron deflection in an electric field.
A practice routine
Redo the worked example with q = −5.0 µC for all four fields.
Find the charge that would make the horizontal-field tilt exactly 45°.
Explain in two sentences why a charge at the pivot leaves the period unchanged.
Solve three past-paper problems on pendulums in fields, writing g_eff as a vector each time.
When you want to see how the tilt and period change together, the Fermi Simulation Lab has interactive simulations for ideas that are easier to see than to read.
Try step 2 now: the answer needs only qE = mg.
