Two infinite line charges are parallel to the z-axis and pass through the x-axis at x= ±d as shown. Find the energy required to move a negative point charge -q from the origin (midway between the two line charges) to a point h along the y-axis. Recall that the field and potential of a single uniform line charge are given by E(r)=f_Pro 2πεor Φ(r)=- where r is the distance from the line charge normal to its axis. ριο 2περ - Inr
Two infinite line charges are parallel to the z-axis and pass through the x-axis at x= ±d as shown. Find the energy required to move a negative point charge -q from the origin (midway between the two line charges) to a point h along the y-axis. Recall that the field and potential of a single uniform line charge are given by E(r)=f_Pro 2πεor Φ(r)=- where r is the distance from the line charge normal to its axis. ριο 2περ - Inr
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![**Problem Description:**
Two infinite line charges are parallel to the z-axis and pass through the x-axis at \( x = \pm d \) as shown. Find the energy required to move a negative point charge \(-q\) from the origin (midway between the two line charges) to a point \( h \) along the y-axis. Recall that the field and potential of a single uniform line charge are given by:
\[
\vec{E}(r) = \hat{r} \frac{\rho_{\ell 0}}{2\pi \varepsilon_0 r}
\]
\[
\Phi(r) = -\frac{\rho_{\ell 0}}{2\pi \varepsilon_0} \ln r
\]
where \( r \) is the distance from the line charge normal to its axis.
**Diagram Explanation:**
The diagram on the right shows two infinite line charges running parallel to the z-axis. These lines are denoted with charge density \(\rho_{\ell 0}\). They are positioned symmetrically at \( x = d \) and \( x = -d \) on the x-axis. The negative point charge \(-q\) is located at the origin, at equal distance from both line charges, and needs to be moved to a point \( h \) along the positive y-axis.
The electrostatic field and potential equations provided are essential for calculating the energy required to move the charge \(-q\) from its initial to final position.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F45e73a8e-4729-402e-8121-de43777babe4%2F83d7c10f-ce06-4984-8368-37db8318522f%2Fj12gqcw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
Two infinite line charges are parallel to the z-axis and pass through the x-axis at \( x = \pm d \) as shown. Find the energy required to move a negative point charge \(-q\) from the origin (midway between the two line charges) to a point \( h \) along the y-axis. Recall that the field and potential of a single uniform line charge are given by:
\[
\vec{E}(r) = \hat{r} \frac{\rho_{\ell 0}}{2\pi \varepsilon_0 r}
\]
\[
\Phi(r) = -\frac{\rho_{\ell 0}}{2\pi \varepsilon_0} \ln r
\]
where \( r \) is the distance from the line charge normal to its axis.
**Diagram Explanation:**
The diagram on the right shows two infinite line charges running parallel to the z-axis. These lines are denoted with charge density \(\rho_{\ell 0}\). They are positioned symmetrically at \( x = d \) and \( x = -d \) on the x-axis. The negative point charge \(-q\) is located at the origin, at equal distance from both line charges, and needs to be moved to a point \( h \) along the positive y-axis.
The electrostatic field and potential equations provided are essential for calculating the energy required to move the charge \(-q\) from its initial to final position.
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