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$_0$.