A proton sists at coordinates (x,y)=(0, 0), and an electron at (d, h), where `d gt gt h`. At time `t=0` a uniform electric field E of unknown magnitude but pointing in the positive y direction is turned on. Assuming that d is large enough that the proton electron interaction is negligible, the y coordinates of the two particles will be equal (at equal time)

Options

A. at about y `=h//2000`
B. at an undetermined value since E is unknown,
C. at about `y=h//43`,
D. nowhere: they move in opposite sirections

Explanation

To solve the problem, we need to analyze the motion of a proton and an electron in a uniform electric field directed in the positive y-direction. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
- The proton is at coordinates (0, 0) and the electron is at (d, h) where \(d \gg h\).
- An electric field \(E\) is turned on at \(t = 0\) in the positive y-direction.

**Hint:** Identify the initial positions of the particles and the direction of the electric field.

### Step 2: Determine the Forces Acting on Each Particle
- The force on the proton (\(F_p\)) due to the electric field is given by:
\[ F_p = q_p E \]
where \(q_p\) is the charge of the proton.

- The force on the electron (\(F_e\)) is:
\[ F_e = q_e E \]
where \(q_e\) is the charge of the electron (note that \(q_e\) is negative).

**Hint:** Recall that the force on a charged particle in an electric field is given by \(F = qE\).

### Step 3: Calculate the Accelerations of Each Particle
- The acceleration of the proton (\(a_p\)) is:
\[ a_p = \frac{F_p}{m_p} = \frac{q_p E}{m_p} \]

- The acceleration of the electron (\(a_e\)) is:
\[ a_e = \frac{F_e}{m_e} = \frac{q_e E}{m_e} \]

**Hint:** Use Newton's second law \(F = ma\) to express the accelerations in terms of the forces and masses.

### Step 4: Write the Equations of Motion
- For the proton, starting from rest, the distance traveled in the y-direction (\(y\)) after time \(t\) is:
\[ y = \frac{1}{2} a_p t^2 = \frac{1}{2} \left(\frac{q_p E}{m_p}\right) t^2 \]

- For the electron, the distance traveled in the y-direction (\(h\)) is:
\[ h' = \frac{1}{2} a_e t^2 = \frac{1}{2} \left(\frac{-q_e E}{m_e}\right) t^2 \]

**Hint:** Remember that the electron moves in the opposite direction due to its negative charge.

### Step 5: Set the Distances Equal
At equal time \(t\), we want to find when the y-coordinates of the proton and electron are equal:
\[ y = h' \]
Substituting the expressions from Step 4:
\[ \frac{1}{2} \left(\frac{q_p E}{m_p}\right) t^2 = \frac{1}{2} \left(\frac{-q_e E}{m_e}\right) t^2 \]

**Hint:** Cancel out common factors and simplify the equation.

### Step 6: Solve for the Ratio of Distances
Cancelling \(t^2\) and \(\frac{1}{2}\) from both sides gives:
\[ \frac{q_p}{m_p} = \frac{-q_e}{m_e} \]
From this, we can express the ratio of the distances traveled by the proton and electron:
\[ \frac{y}{h} = \frac{m_e}{m_p} \]

**Hint:** Use the known mass ratio of the proton and electron to simplify further.

### Step 7: Substitute Known Mass Values
The mass of the proton \(m_p\) is approximately 1836 times the mass of the electron \(m_e\):
\[ \frac{y}{h} = \frac{m_e}{1836 m_e} = \frac{1}{1836} \]
Thus, we find:
\[ y = \frac{h}{1836} \]

**Hint:** Ensure you understand how the mass ratio affects the final distance traveled.

### Conclusion
The y-coordinate of the proton will be equal to \(\frac{h}{1836}\) when the y-coordinate of the electron is \(h\).

### Final Answer
The y-coordinate of the proton when it equals the y-coordinate of the electron is:
\[ y = \frac{h}{1836} \]