The electric potential of a system is presented by the figure shown below. Find the x-component of the electric field at points A, B, and C if V, = 12 V and x, = 25 cm, x2 = 100 cm, x3 = 150 cm. В V. А C X1 X2 X3 The x-component of E-field at point A, E = 48 XUnits N/C The x-component of E-field at point B, E, =0 VUnits N/C The x-component of E-field at point C, Ec = -24 X Units N/C
The electric potential of a system is presented by the figure shown below. Find the x-component of the electric field at points A, B, and C if V, = 12 V and x, = 25 cm, x2 = 100 cm, x3 = 150 cm. В V. А C X1 X2 X3 The x-component of E-field at point A, E = 48 XUnits N/C The x-component of E-field at point B, E, =0 VUnits N/C The x-component of E-field at point C, Ec = -24 X Units N/C
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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
Transcribed Image Text:The figure shows a graph of electric potential (V) versus position (x), illustrating a system with three key points: A, B, and C. The potential values and specific x-coordinates are given to find the x-component of the electric field at these points.
**Graph Details:**
- **Axes:**
- The vertical axis represents the electric potential (V) with a labeled value \( V_0 \).
- The horizontal axis represents the position (x).
- **Points:**
- **A**: Located at \( x_1 \) where the potential is \( V_0 \).
- **B**: Between \( x_1 \) and \( x_2 \) where the potential remains constant at \( V_0 \).
- **C**: At \( x_3 \) where the potential drops to zero.
**x-Coordinates and Potential:**
- \( x_1 = 25 \, \text{cm} \)
- \( x_2 = 100 \, \text{cm} \)
- \( x_3 = 150 \, \text{cm} \)
- \( V_0 = 12 \, \text{V} \)
**Calculation of the Electric Field (E):**
The electric field is calculated as the negative gradient (slope) of the potential vs. position graph.
- **E-field at point A (\( E_A \)):**
\( E_A = -\frac{dV}{dx} \) over the segment 0 to \( x_1 \)
- Computed value: \( -48 \, \text{N/C} \) (incorrect value indicated with red 'X')
- **E-field at point B (\( E_B \)):**
No change in potential between \( x_1 \) and \( x_2 \), so \( E_B = 0 \, \text{N/C} \)
- **E-field at point C (\( E_C \)):**
\( E_C = -\frac{dV}{dx} \) over the segment \( x_2 \) to \( x_3 \)
- Computed value: \( -24 \, \text{N/C} \)
This educational example demonstrates how to derive electric field components from a potential vs. position graph.
![A particle of mass 5 g and charge 21.5 µC is moving in an electric potential field
\[ V(x, y) = c_1 \cdot x - c_2 \cdot y^2 + c_3 \cdot y \cdot x^2, \]
where \( c_1 = 85 \, \text{V/m}, \, c_2 = 10 \, \text{V/m}^2, \) and \( c_3 = 55 \, \text{V/m}^3. \) Find the electric field acting on the particle as a function of its position. Use V/m and meters for the units, but do not put them explicitly in \(\mathbf{E}(x, y)\).
- The x-component of the E-field, \( E_x(x,y) = -55 \) V/m.
- The y-component of the E-field, \( E_y(x,y) = 20 \) V/m.
What is the magnitude of the particle's acceleration at \( x = 1 \, \text{m} \) and \( y = -0.5 \, \text{m} \)?
- The acceleration, \( a = 0.3077 \) \(\text{m/s}^2\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fabdaa8be-76ad-4614-bb55-5dc3ce3d6a6b%2F545f4437-d896-410e-b0d2-3e5e10753ffd%2Fa5sj0sa_processed.png&w=3840&q=75)
Transcribed Image Text:A particle of mass 5 g and charge 21.5 µC is moving in an electric potential field
\[ V(x, y) = c_1 \cdot x - c_2 \cdot y^2 + c_3 \cdot y \cdot x^2, \]
where \( c_1 = 85 \, \text{V/m}, \, c_2 = 10 \, \text{V/m}^2, \) and \( c_3 = 55 \, \text{V/m}^3. \) Find the electric field acting on the particle as a function of its position. Use V/m and meters for the units, but do not put them explicitly in \(\mathbf{E}(x, y)\).
- The x-component of the E-field, \( E_x(x,y) = -55 \) V/m.
- The y-component of the E-field, \( E_y(x,y) = 20 \) V/m.
What is the magnitude of the particle's acceleration at \( x = 1 \, \text{m} \) and \( y = -0.5 \, \text{m} \)?
- The acceleration, \( a = 0.3077 \) \(\text{m/s}^2\).
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