Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC volta

Q: Why is the net force at $45^\circ$?

Because the proton moves at $45^\circ$ from rest, its acceleration must be at $45^\circ$, which means the net force is at $45^\circ$.

Q: Why do we equate $qE$ and $mg$?

Since the net force is at $45^\circ$ to the vertical, the horizontal and vertical components of the force must be equal. $\tan 45^\circ = 1$.

Q: What is the formula for electric field between parallel plates?

The electric field $E$ is uniform and is given by $E = \frac{V}{d}$, where $V$ is the potential difference and $d$ is the separation.

Q: Does the position of the proton between the plates matter?

No, the electric field between large parallel plates is uniform, so the force $qE$ is the same everywhere between the plates.

Q: Why did we convert $1\text{ cm}$ to meters?

To ensure all units are in the standard SI system before calculation, preventing order-of-magnitude errors.

Question

Two large vertical and parallel metal plates having a separation of $1	ext{ cm}$ are connected to a DC voltage source of potential difference $X$. A proton is released at rest midway between the two plates. It is found to move at $45^\circ$ to the vertical just after release. Then $X$ is nearly

Options:

(A) $1 	imes 10^{-5}	ext{ V}$

(B) $1 	imes 10^{-7}	ext{ V}$

(C) $1 	imes 10^{-9}	ext{ V}$

(D) $1 	imes 10^{-10}	ext{ V}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing between two massive, vertical metal plates. These plates are connected to a DC voltage source, creating a uniform electric field between them. Right in the middle of this space, a tiny proton is released from rest.

What happens next? The proton doesn't just fall straight down due to gravity, nor does it move purely horizontally due to the electric field. Instead, it moves diagonally, exactly at a $45^\circ$ angle to the vertical. This specific angle is the key to unlocking the entire problem.

### The Master Equation

When a particle is released from rest, its initial velocity is zero. The direction of its motion is entirely dictated by the direction of the net force acting on it.

There are two primary forces at play here:
1. The downward gravitational force: $F_g = mg$
2. The horizontal electrostatic force: $F_e = qE$

Since the proton moves at a $45^\circ$ angle to the vertical, the horizontal and vertical components of the net force must be perfectly balanced. Mathematically, we can express this using trigonometry:
$$ 	an 45^\circ = \frac{F_e}{F_g} $$

Because $	an 45^\circ = 1$, we arrive at a beautifully simple relationship:
$$ F_e = F_g $$
$$ qE = mg $$

We also know that the electric field $E$ between two parallel plates separated by a distance $d$ with a potential difference $X$ is given by:
$$ E = \frac{X}{d} $$

Substituting this into our force balance equation, we get:
$$ q\left(\frac{X}{d}ight) = mg $$

### Final Calculation

Our goal is to find the potential difference $X$. Let's rearrange the equation to isolate $X$:
$$ X = \frac{mgd}{q} $$

Now, it's time to plug in the fundamental constants and the given values. 
- Mass of a proton, $m = 1.67 	imes 10^{-27}	ext{ kg}$
- Acceleration due to gravity, $g = 9.8	ext{ m/s}^2$
- Separation distance, $d = 1	ext{ cm} = 10^{-2}	ext{ m}$
- Charge of a proton, $q = 1.6 	imes 10^{-19}	ext{ C}$

$$ X = \frac{(1.67 	imes 10^{-27})(9.8)(10^{-2})}{1.6 	imes 10^{-19}} $$

Let's carefully compute the numerator first:
$$ 1.67 	imes 9.8 \approx 16.366 $$
$$ 16.366 	imes 10^{-29} $$

Now, divide by the charge:
$$ X = \frac{16.366 	imes 10^{-29}}{1.6 	imes 10^{-19}} \approx 10.22 	imes 10^{-10}	ext{ V} $$

This can be rewritten in standard scientific notation as:
$$ X \approx 1.022 	imes 10^{-9}	ext{ V} $$

Looking at our options, this is nearly $1 	imes 10^{-9}	ext{ V}$. The elegance of this problem lies in how a simple geometric observation—a $45^\circ$ angle—allows us to directly equate two fundamentally different forces of nature!

Select Answer:

Visualized Solution

Forces on the Proton

Condition for $4 5^{\circ}$ Motion

Equating the Forces

Rearranging for $X$

Substituting Values

Final Calculation

The Way Forward

The Sigma Insight: Electric Field

Analyzing the Setup

The Master Equation

Final Calculation

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Select Answer:

Visualized Solution

Forces on the Proton

Condition for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 45∘ Motion

Equating the Forces

Rearranging for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } X

Substituting Values

Final Calculation

The Way Forward

The Sigma Insight: Electric Field

Analyzing the Setup

The Master Equation

Final Calculation

Similar Questions

Condition for $4 5^{\circ}$ Motion

Rearranging for $X$