If `g_E` and `g_M` are the acceleration due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio
electronic charge on the moon/electronic charge on the earth to be

Options

A. (a) `g_M//g_E`
B. (b) 1
C. (c) 0
D. (d) `g_E//g_M`

Explanation

To solve the problem, we need to understand the relationship between the electronic charge and the acceleration due to gravity on the Earth and the Moon. The key point is that the electronic charge (denoted as \( e \)) is a fundamental constant and does not depend on the local gravitational field strength.

### Step-by-Step Solution:

1. **Understanding the Experiment**: Millikan's oil drop experiment is designed to measure the charge of an electron. The experiment involves balancing the gravitational force acting on an oil drop with an electric force.

2. **Gravitational Forces**: The gravitational force acting on an oil drop can be expressed as:
\[ F_g = m \cdot g \]
where \( m \) is the mass of the oil drop and \( g \) is the acceleration due to gravity (which is different on the Earth and the Moon).

3. **Electric Force**: The electric force acting on the charged oil drop is given by:
\[ F_e = q \cdot E \]
where \( q \) is the charge on the oil drop (which we are trying to measure) and \( E \) is the electric field strength.

4. **Balancing Forces**: In Millikan's experiment, the oil drop reaches a terminal velocity when the gravitational force is balanced by the electric force:
\[ m \cdot g = q \cdot E \]

5. **Ratio of Charges**: The charge \( q \) does not depend on the gravitational field strength. Therefore, when performing the experiment on the Moon, the charge measured will still be the same as that measured on Earth.

6. **Conclusion**: Since the electronic charge does not depend on the local gravitational field, the ratio of the electronic charge on the Moon to the electronic charge on the Earth is:
\[ \frac{q_{\text{Moon}}}{q_{\text{Earth}}} = 1 \]

### Final Answer:
The ratio of the electronic charge on the Moon to the electronic charge on the Earth is \( 1 \).

---