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The problem consists of a setup involving two parallel plates separated by a distance of 20 cm, with a potential difference between them of -2V on the left plate and +10V on the right plate.

Let's break down the tasks in the question:

### Part (a)
1. **Given:** The potential difference varies linearly across the plates.
2. **Task:** Find the equation for the potential \( V(x) \) as a function of distance \( x \), where \( x = 0 \) corresponds to the -2V plate and \( x = 0.2 \) meters (20 cm) corresponds to the +10V plate.

The general form for a linearly varying potential is:
\[
V(x) = mx + c
\]
where \( m \) is the slope and \( c \) is the y-intercept.

Since the change in voltage is linear:
- The difference in potential between the plates is \( 10 - (-2) = 12 \) volts.
- The distance between the plates is \( 0.2 \) meters.

The slope \( m \) can be calculated as:
\[
m = \frac{V_{\text{final}} - V_{\text{initial}}}{x_{\text{final}} - x_{\text{initial}}} = \frac{10 - (-2)}{0.2 - 0} = \frac{12}{0.2} = 60 \, \text{V/m}
\]
Thus, the equation becomes:
\[
V(x) = 60x - 2
\]

### Part (b)
1. **Given:** A particle with mass \( m = 1 \, \text{g} \) and charge \( q = -2 \, \text{C} \) is located at the -2V plate initially.
2. **Task:** Calculate the force acting on the particle, assuming it remains stationary, and then find its velocity after traveling a distance of 5 cm (0.05 m) from the -2V plate.

To solve part (b):
1. **Electric Field (E):** Since the potential varies linearly, the electric field \( E \) is constant and given by:
   \[
   E = -\frac{dV}{dx} = -60 \, \text{V/m}
   \]

2. **Force (F):** The force on a charge in an electric field is given by:
   \[
   F = qE = (-2) \times (-60) = 120 \, \text{N}
   \]

3. **Acceleration (a):** Using Newton's second law \( F = ma \), we find the acceleration:
   \[
   a = \frac{F}{m} = \frac{120}{0.001} = 120000 \, \text{m/s}^2
   \]

4. **Velocity after traveling 5 cm:** Assuming it starts from rest, we use \( v^2 = u^2 + 2as \):
   \[
   v = \sqrt{0 + 2 \times 120000 \times 0.05} = \sqrt{12000} \approx 109.54 \, \text{m/s}
   \]

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Would you like a more detailed breakdown of any steps or further clarification? 

### Related Questions:
1. How does the mass of the particle affect the final velocity?
2. What would happen if the charge were positive instead of negative?
3. How does the distance between the plates influence the electric field strength?
4. If the particle had an initial velocity, how would that change the result?
5. What is the effect of a non-linear potential distribution on the electric field?

### Tip:
In problems involving a linearly varying electric potential, remember that the electric field will be constant and equal to the slope of the potential function.

در شکل بالا اگر پتانسیل بطور خطی تغییر کند الف) معادله تغییرات پتانسیل مابین دو صفحه را پیدا کنید. ب) اگر ذره ای به جرم یک گرم و بار -2 کولن در صفحه ای -2 ولتی ساکن باشد سرعت ذره را در فاصله 5 سانتی متری از این صفحه محاسبه کنید.

Explore the solution to a physics problem involving a linearly varying electric potential between two plates with -2V and 10V potentials separated by 20 cm. Calculate the potential equation, the electric field, and the velocity of a charged particle with mass 1 g and charge -2 C after moving 5 cm. Suitable for undergraduate physics students.

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