A simple pendulum has a bob of mass m carrying a positive charge q. It is placed in a region where a uniform electric field E is directed vertically upwards. If the string is displaced slightly and released, what happens to its time period T compared to its time period T$_0$ without the electric field?
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Remember how various forces affect effective gravity:
- Upward force (electric, buoyant, etc.) reduces $g_{eff}$, thus increasing the time period.
- Downward force increases $g_{eff}$, thus decreasing the time period.
- Free fall means $g_{eff} = 0$, leading to infinite time period (no oscillation).
The period of a simple pendulum is \( T = 2\pi\sqrt{\frac{L}{g_{eff}}} \), so \( T \) grows whenever the effective gravity \( g_{eff} \) shrinks.
T \textgreater T\(_0\): The upward electric force \( qE \) on the positively charged bob works against gravity, so the net downward force, and hence \( g_{eff} \), is smaller than before; a smaller \( g_{eff} \) means a longer period.
T \textless T\(_0\): Would need \( g_{eff} \) to increase, which would require an additional downward force, not the upward one described here.
T = T\(_0\): Would need no change in \( g_{eff} \) at all, which is not the case since a nonzero upward electric force is acting.
Will not oscillate: Would need the upward electric force to be as large as or larger than gravity itself, which is not indicated by the problem.
Therefore, the correct answer is T \textgreater T\(_0\).