Question
What is the electric potential at a point on the x-axis due to a charge distributed along the x-axis from −1 m to 1 m?
Answers
GPT-4o mini
To find the electric potential \( V \) at a point on the x-axis due to a charge distributed along the x-axis from \( -1 \) m to \( 1 \) m, we need to consider the contributions to the potential from each infinitesimal segment of charge along the distribution.
### Step-by-Step Calculation
1. **Setup the Problem**:
- Let's denote the linear charge density as \( \lambda \) (charge per unit length).
- Consider a point \( P \) on the x-axis at position \( x \) where we want to calculate the electric potential.
- The charge distribution is along the segment from \( x = -1 \) m to \( x = 1 \) m.
2. **Expression for Potential**:
The electric potential \( dV \) due to an infinitesimal charge \( dq \) located at position \( x' \) on the x-axis is given by:
\[
dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r}
\]
where \( r \) is the distance from the charge \( dq \) to the point \( P \).
3. **Infinitesimal Charge Element**:
Let \( dx' \) be an infinitesimal length element of the charge distribution, then:
\[
dq = \lambda \, dx'
\]
4. **Distance Calculation**:
The distance \( r \) from the point \( P \) located at \( x \) to the charge at position \( x' \) is:
\[
r = |x - x'|
\]
5. **Integrating for Total Potential**:
The potential at point \( x \) due to the entire charge distribution can be found by integrating \( dV \) from \( x' = -1 \) to \( x' = 1 \):
\[
V(x) = \int_{-1}^{1} dV = \int_{-1}^{1} \frac{1}{4\pi\epsilon_0} \frac{\lambda \, dx'}{|x - x'|}
\]
This simplifies to:
\[
V(x) = \frac{\lambda}{4\pi\epsilon_0} \int_{-1}^{1} \frac{dx'}{|x - x'|}
\]
6. **Evaluate the Integral**:
The integral can be evaluated differently depending on the position of \( x \) relative to the interval \([-1, 1]\):
- **Case 1**: For \( x < -1 \) or \( x > 1 \), the potential can be computed by considering the limits separately, since \( |x - x'| = -(x - x') \) when \( x < -1 \) and \( |x - x'| = x - x' \) when \( x > 1\).
- **Case 2**: For \( -1 < x < 1 \), we will have to handle it differently. Here, the integral can be split at \( x \) leading to different contributions.
### Final Expression
When evaluated, the result should yield a formula that captures dependence on \( x \) and constants regarding \( \lambda \) and \( \epsilon_0 \).
### Conclusion
To sum up, the electric potential \( V(x) \) at a point \( x \) on the x-axis due to a uniformly distributed charge along the segment from \( -1 \) to \( 1 \) is given by:
\[
V(x) = \frac{\lambda}{4\pi\epsilon_0} \int_{-1}^{1} \frac{1}{|x - x'|} \, dx'
\]
You would then evaluate this integral depending on the desired region of interest (not forgetting the physical interpretation if \( x \) is within, outside, or at bounds of the charge distribution).
### Step-by-Step Calculation
1. **Setup the Problem**:
- Let's denote the linear charge density as \( \lambda \) (charge per unit length).
- Consider a point \( P \) on the x-axis at position \( x \) where we want to calculate the electric potential.
- The charge distribution is along the segment from \( x = -1 \) m to \( x = 1 \) m.
2. **Expression for Potential**:
The electric potential \( dV \) due to an infinitesimal charge \( dq \) located at position \( x' \) on the x-axis is given by:
\[
dV = \frac{1}{4\pi\epsilon_0} \frac{dq}{r}
\]
where \( r \) is the distance from the charge \( dq \) to the point \( P \).
3. **Infinitesimal Charge Element**:
Let \( dx' \) be an infinitesimal length element of the charge distribution, then:
\[
dq = \lambda \, dx'
\]
4. **Distance Calculation**:
The distance \( r \) from the point \( P \) located at \( x \) to the charge at position \( x' \) is:
\[
r = |x - x'|
\]
5. **Integrating for Total Potential**:
The potential at point \( x \) due to the entire charge distribution can be found by integrating \( dV \) from \( x' = -1 \) to \( x' = 1 \):
\[
V(x) = \int_{-1}^{1} dV = \int_{-1}^{1} \frac{1}{4\pi\epsilon_0} \frac{\lambda \, dx'}{|x - x'|}
\]
This simplifies to:
\[
V(x) = \frac{\lambda}{4\pi\epsilon_0} \int_{-1}^{1} \frac{dx'}{|x - x'|}
\]
6. **Evaluate the Integral**:
The integral can be evaluated differently depending on the position of \( x \) relative to the interval \([-1, 1]\):
- **Case 1**: For \( x < -1 \) or \( x > 1 \), the potential can be computed by considering the limits separately, since \( |x - x'| = -(x - x') \) when \( x < -1 \) and \( |x - x'| = x - x' \) when \( x > 1\).
- **Case 2**: For \( -1 < x < 1 \), we will have to handle it differently. Here, the integral can be split at \( x \) leading to different contributions.
### Final Expression
When evaluated, the result should yield a formula that captures dependence on \( x \) and constants regarding \( \lambda \) and \( \epsilon_0 \).
### Conclusion
To sum up, the electric potential \( V(x) \) at a point \( x \) on the x-axis due to a uniformly distributed charge along the segment from \( -1 \) to \( 1 \) is given by:
\[
V(x) = \frac{\lambda}{4\pi\epsilon_0} \int_{-1}^{1} \frac{1}{|x - x'|} \, dx'
\]
You would then evaluate this integral depending on the desired region of interest (not forgetting the physical interpretation if \( x \) is within, outside, or at bounds of the charge distribution).