Chapter 21 Problem 31

Problem

${\vec {E}}_{\textrm {thread}}$ and ${\vec {E}}_{\textrm {Q}}$ represent fields due to the long thread and the charge Q, respectively

A long uniformly charged thread (linear charge density $\lambda =2.5{\textrm {C/m}}$ ) lies along the $x$ axis. A small charged sphere ($Q=-2.0{\textrm {C}}$) is at the point $x=0{\textrm {cm}}$, $y=-5.0{\textrm {cm}}$. What is the electric field at the point $x=7.0{\textrm {cm}}$, $y=7.0{\textrm {cm}}$?

Solution

${\vec {E}}_{\textrm {thread}}={\frac {1}{2\pi \epsilon _{0}}}{\frac {\lambda }{y}}{\hat {j}}$

${\vec {E}}_{\textrm {Q}}={\frac {1}{4\pi \epsilon _{0}}}{\frac {Q}{d^{2}}}\left(-\cos \theta {\hat {i}}-\sin \theta {\hat {j}}\right)$

${\vec {E}}=\left(-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Q}{d^{2}}}\cos \theta \right){\hat {i}}+\left({\frac {1}{2\pi \epsilon _{0}}}{\frac {\lambda }{y}}-{\frac {1}{4\pi \epsilon _{0}}}{\frac {Q}{d^{2}}}\sin \theta \right){\hat {j}}$

$d^{2}=(0.07{\textrm {m}})^{2}+(0.12{\textrm {m}})^{2}=0.0193{\textrm {m}}^{2}$

$\theta =\tan ^{-1}{\frac {12.0{\textrm {cm}}}{7.0{\textrm {cm}}}}=59.7^{\circ }$

$E_{x}={\vec {E}}{\dot {\hat {i}}}=-4.699\times 10^{11}{\textrm {N/C}}$ $E_{y}={\vec {E}}{\dot {\hat {j}}}=-1.622\times 10^{11}{\textrm {N/C}}$

${\vec {E}}=\left(-4.699{\hat {i}}-1.6{\hat {j}}\right)\times 10^{11}{\textrm {N/C}}$

$E=\left|{\vec {E}}\right|={\sqrt {E_{x}^{2}+E_{y}^{2}}}=5.0\times 10^{11}{\textrm {N/C}}$

$\theta _{E}=\tan ^{-1}{\frac {-1.622}{-4.699}}=199^{\circ }$