Two insulating plates are both uniformly charged in such a way that the potential difference between

Q: Why does the electron move towards Plate 2 if the electric field points towards Plate 1?

Great question! Electric field lines show the direction of force on a *positive* charge. Since the electron is *negatively* charged, it experiences a force in the exact opposite direction of the electric field.

Q: Why didn't we use the distance $d = 0.1\text{ m}$ in our calculation?

The distance is actually a distractor! The work done by an electric field on a charge depends only on the potential difference between the start and end points ($W = q\Delta V$), regardless of the distance between them.

Q: Can I solve this using Newton's laws of motion instead of energy?

Absolutely! You can find the electric field ($E = V/d$), then the acceleration ($a = qE/m$), and finally use $v^2 = u^2 + 2ad$. However, the Work-Energy theorem is much faster and less prone to calculation errors.

Q: Why is the initial kinetic energy zero?

The problem explicitly states that the electron is 'released from rest'. This means its initial velocity $u = 0$, so its initial kinetic energy is also zero.

Q: What would happen if we released a proton instead of an electron?

A proton has a positive charge, so it would experience a force in the *same* direction as the electric field (towards Plate 1). If released at Plate 1, it wouldn't move towards Plate 2 at all!

Question

Two insulating plates are both uniformly charged in such a way that the potential difference between them is $V_2 - V_1 = 20	ext{ V}$ (i.e., plate 2 is at a higher potential). The plates are separated by $d = 0.1	ext{ m}$ and can be treated as infinitely large. An electron is released from rest on the inner surface of plate 1. What is its speed when it hits plate 2? ($e = 1.6 	imes 10^{-19}	ext{ C}, m_e = 9.11 	imes 10^{-31}	ext{ kg}$)

Options:

(A) $2.65 	imes 10^6	ext{ ms}^{-1}$

(B) $7.02 	imes 10^{12}	ext{ ms}^{-1}$

(C) $1.87 	imes 10^6	ext{ ms}^{-1}$

(D) $32 	imes 10^{-19}	ext{ ms}^{-1}$

Sigma Percentile · Accepted Answer

### Visualizing the Setup

Imagine two infinitely large, parallel insulating plates standing vertically. We are told that Plate 2 is at a higher potential than Plate 1, with a potential difference of $V_2 - V_1 = 20	ext{ V}$. The plates are separated by a distance $d = 0.1	ext{ m}$.

Because Plate 2 is at a higher potential, the electric field $\vec{E}$ in the region between the plates must point from Plate 2 towards Plate 1. Remember, electric field lines always flow from regions of higher potential to lower potential, much like water flowing down a hill.

### The Invisible Hand: Electric Field and Force

Now, we release an electron from rest at the inner surface of Plate 1. An electron carries a negative charge ($-e$). According to the laws of electrostatics, a negative charge experiences an electric force $\vec{F}_e$ in the direction *opposite* to the electric field.

Since the electric field points to the left (towards Plate 1), the electron is pushed to the right (towards Plate 2). It starts from rest ($u = 0$) and accelerates across the gap.

### The Work-Energy Theorem

To find the final speed $v$ of the electron just before it crashes into Plate 2, we could use kinematics, but there is a much more elegant tool: the **Work-Energy Theorem**. This theorem states that the work done by the net force on an object equals its change in kinetic energy.

$$W = \Delta K$$

The work done by the electric field on a charge $q$ moving through a potential difference $\Delta V$ is given by $W = q\Delta V$. For our electron, the charge is $e$ and it moves through a potential difference of $V_2 - V_1$. The change in kinetic energy is simply its final kinetic energy minus its initial kinetic energy (which is zero).

$$\frac{1}{2}m_e v^2 = e(V_2 - V_1)$$

Notice something fascinating here? The distance $d = 0.1	ext{ m}$ doesn't even appear in our equation! The final speed depends entirely on the potential difference, not how far apart the plates are.

### Crunching the Numbers

Now, it's just a matter of plugging in the constants. We know the mass of an electron $m_e = 9.11 	imes 10^{-31}	ext{ kg}$, the elementary charge $e = 1.6 	imes 10^{-19}	ext{ C}$, and the potential difference is $20	ext{ V}$.

$$\frac{1}{2}(9.11 	imes 10^{-31}) v^2 = (1.6 	imes 10^{-19})(20)$$

Let's isolate $v^2$ by multiplying both sides by 2 and dividing by the mass:

$$v^2 = \frac{1.6 	imes 10^{-19} 	imes 40}{9.11 	imes 10^{-31}}$$

$$v^2 = \frac{64 	imes 10^{-19}}{9.11 	imes 10^{-31}} = \frac{64}{9.11} 	imes 10^{12}$$

Taking the square root of both sides yields the final speed:

$$v \approx 2.65 	imes 10^6	ext{ ms}^{-1}$$

The electron reaches a staggering speed of over 2.6 million meters per second! This beautifully demonstrates how even a small potential difference of 20 volts can impart massive velocities to subatomic particles due to their incredibly tiny mass.

Select Answer:

Visualized Solution

Visualizing the Setup

Electric Field Direction

Force on the Electron

Work-Energy Theorem

Setting up the Equation

Substituting Values

Isolating Velocity

Final Calculation

The Way Forward

The Sigma Insight: Electric Potential and Potential Difference

Visualizing the Setup

The Invisible Hand

The Work-Energy Theorem

Crunching the Numbers

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Select Answer:

Visualized Solution

Visualizing the Setup

Electric Field Direction

Force on the Electron

Work-Energy Theorem

Setting up the Equation

Substituting Values

Isolating Velocity

Final Calculation

The Way Forward

The Sigma Insight: Electric Potential and Potential Difference

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