Question

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?

• A. T \textgreater T$_0$
• B. T \textless T$_0$
• C. T = T$_0$
• D. The pendulum will not oscillate

Hint:

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).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks how the time period of a simple pendulum changes when it is placed in a uniform upward electric field, compared to its time period without the field. The pendulum bob has a positive charge.

Step 2: Key Formula or Approach:
The time period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g_{eff}}} \]
Where:
- \( L \) is the length of the pendulum.
- \( g_{eff} \) is the effective acceleration due to gravity.

Step 3: Detailed Explanation:
1. Without the electric field:
The only force acting downwards is gravity, $mg$.
The effective acceleration due to gravity is $g_{eff} = g$.
The time period is $T_0 = 2\pi \sqrt{\frac{L}{g}}$.
2. With the electric field:
The bob has mass $m$ and positive charge $q$.
The electric field $E$ is directed vertically upwards.
* Gravitational force: $F_g = mg$ (acting downwards).
* Electric force: $F_e = qE$ (acting upwards, since $q$ is positive and $E$ is upwards).
The net downward force is $F_{net} = mg - qE$.
The effective acceleration due to gravity is $g'_{eff} = \frac{F_{net}}{m} = \frac{mg - qE}{m} = g - \frac{qE}{m}$.
3. Comparison of time periods:
Assuming the pendulum still oscillates (i.e., $mg > qE$, so $g'_{eff} > 0$):
Since $g'_{eff} = g - \frac{qE}{m}$, and $\frac{qE}{m}$ is a positive quantity, it implies $g'_{eff} < g$.
The new time period is $T = 2\pi \sqrt{\frac{L}{g'_{eff}}}$.
Since $g'_{eff}$ is smaller than $g$, the term $\frac{1}{g'_{eff}}$ will be larger than $\frac{1}{g}$.
Therefore, $T > T_0$.

Step 4: Final Answer:
The new time period T is greater than T$_0$.

Answer 2

Rather than deriving the effective gravity formula again, this approach checks what condition on the forces would have to hold for each of the four possible outcomes, and compares that against the situation actually described.

1. T \textgreater T\(_0\): For the period to increase, the net restoring force on the bob for a given displacement must be weaker than it would be under gravity alone, which happens when an upward force partially cancels gravity. Here, the field \( E \) points upward and the bob's charge \( q \) is positive, so the electric force \( qE \) acts upward, opposing and reducing the net downward pull. This directly weakens the restoring force, so the period should indeed increase, matching this option.
2. T \textless T\(_0\): A shorter period would need the net downward force to be stronger than gravity alone, which would require an additional downward force, not an upward one; since the electric force here acts upward, this scenario does not apply.
3. T = T\(_0\): The period would stay the same only if the electric force were zero, which would require either \( q = 0 \) or \( E = 0 \); since the bob does carry a nonzero positive charge and the field is explicitly present, this condition is not met.
4. The pendulum will not oscillate: This outcome would need the upward electric force to equal or exceed the downward gravitational force, i.e. \( qE \geq mg \), leaving no net restoring force to bring the bob back to equilibrium; the problem gives no indication that the field is anywhere near strong enough to fully cancel gravity, so this extreme case does not apply here.

Since the upward electric force only partially reduces the net downward pull without cancelling it, the pendulum keeps oscillating but more slowly, meaning its period increases compared to the field-free case.

Therefore, the correct answer is T \textgreater T\(_0\).