Question

A pendulum consists of a small mass 'm' suspended by a thread of length 'l'. The mass carries a +q charge. The pendulum is placed in uniform electric field 'E' directed vertically upwards. What will be the time period of the oscillation?

Answer 1

Period: `qquad 2 pi sqrt(l/(g - (qE)/m))`

Explanation:

Absent the E-field, period for small angle oscillation would be usual relation:

• `T = 2 pi sqrt(l/g)`

The effect of the E-field, which pushed the mass upwards against the g-field, is to reduce the weight of the mass:

Previously:

• `W = mg`

Now:

• `W' = mg - qE = mg' qquad star`

It would be the same as if gravity itself was adjusted from `g to g'`.

From `star`:

• `g' = (mg - qE)/m = g - (qE)/m`

And so:

`T' = 2 pi sqrt(l/(g')) = 2 pi sqrt(l/(g - (qE)/m))`

Note that if `g - (qE)/m lt 0`, the mass is floating because:

• `g - (qE)/m lt 0 implies qE gt mg`

So the formula for `T'` breaks for understandable reasons