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Problem Analysis

1. Setup: We have two plates separated by a distance of 20 cm, with potentials of -2 V and 10 V, respectively. The electric potential between them changes linearly.

2. Question Parts:

   • Part (a): Find the equation of the potential $V$ as a function of distance $x$ from the plate with potential -2 V.
   • Part (b): A particle with mass 1 gram and charge -2 µC (microcoulombs) starts from the -2 V plate. We need to calculate its speed when it reaches a point 5 cm away from the starting plate.

Solution Steps

Part (a): Equation for Potential Variation

Since the potential changes linearly between -2 V and 10 V over a distance of 20 cm, we can use the concept of a linear gradient.

1. Slope Calculation: $\text{slope} = \frac{\Delta V}{\Delta x} = \frac{10 - (-2)}{20} = \frac{12}{20} = 0.6 \, \text{V/cm}$

2. Equation of Potential: Let $x$ be the distance from the -2 V plate. The potential $V(x)$ at a distance $x$ is: $V(x) = -2 + 0.6 \cdot x$

Part (b): Speed of the Particle at 5 cm

1. Potential Difference: At $x = 5$ cm, calculate the potential: $V(5) = -2 + 0.6 \cdot 5 = -2 + 3 = 1 \, \text{V}$ The particle moves from -2 V to 1 V, resulting in a potential difference $\Delta V = 1 - (-2) = 3$ V.

2. Energy Calculation: The work done on the particle (or change in electric potential energy) is given by: $\Delta U = q \cdot \Delta V$ where $q = -2 \, \mu \text{C} = -2 \times 10^{-6} \, \text{C}$. Thus, $\Delta U = (-2 \times 10^{-6}) \cdot 3 = -6 \times 10^{-6} \, \text{J}$

3. Kinetic Energy and Speed: Since the particle starts from rest, its gain in kinetic energy $\Delta K$ equals the magnitude of $\Delta U$: $\Delta K = \frac{1}{2} m v^2 = 6 \times 10^{-6} \, \text{J}$ where $m = 1 \, \text{g} = 0.001 \, \text{kg}$. Solving for $v$: $v = \sqrt{\frac{2 \cdot 6 \times 10^{-6}}{0.001}} = \sqrt{0.012} \approx 0.1095 \, \text{m/s}$

Final Answers

• (a) The potential as a function of distance $x$: $V(x) = -2 + 0.6 \cdot x$.
• (b) The speed of the particle at 5 cm from the -2 V plate is approximately $0.1095 \, \text{m/s}$.

Would you like more details on any of these steps, or have any questions?

Related Questions

1. What would be the speed of a particle with double the charge but the same mass in this scenario?
2. How would the potential equation change if the distance between the plates were doubled?
3. What is the electric field between the plates based on the linear potential change?
4. How would the speed change if the potential difference were halved?
5. How would the answer change if the particle had a positive charge instead?

Tip

Remember, the electric field between two plates with a linear potential difference can be calculated as the rate of potential change over distance, making it uniform in this case.