Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 17 - Line and Surface Integrals - 17.2 Line Integrals - Exercises - Page 934: 49

Answer

The electric potential at $P$ is $V\left( P \right) = - 10096.7$ volts.

Work Step by Step

The charge distribution can be parametrized by ${\bf{r}}\left( t \right) = \left( {0,t,0} \right)$, ${\ \ \ }$ for $1 \le t \le 3$ Thus, $||{\bf{r}}'\left( t \right)|| = \sqrt {\left( {0,1,0} \right)\cdot\left( {0,1,0} \right)} = 1$ $\delta \left( {{\bf{r}}\left( t \right)} \right) = - t$ The distance ${d_P}$ from $P = \left( {2,0,2} \right)$ to a point $\left( {0,y,0} \right)$ on the charge distribution is ${d_P} = \sqrt {{{\left( { - 2} \right)}^2} + {y^2} + {{\left( { - 2} \right)}^2}} = \sqrt {8 + {y^2}} $ Using Eq. (6), the electric potential at $P$ is given by $V\left( P \right) = k\mathop \smallint \limits_C^{} \frac{{\delta \left( {x,y,z} \right){\rm{d}}s}}{{{d_P}}} = k\mathop \smallint \limits_1^3 \frac{{\delta \left( {x,y,z} \right)}}{{\sqrt {8 + {y^2}} }}{\rm{d}}s$ Using Eq. (4): $\mathop \smallint \limits_C^{} f\left( {x,y,z} \right){\rm{d}}s = \mathop \smallint \limits_a^b f\left( {{\bf{r}}\left( t \right)} \right)||{\bf{r}}'\left( t \right)||{\rm{d}}t$, where in this case $f\left( {x,y,z} \right) = \delta \left( {x,y,z} \right) = - y$, we obtain $V\left( P \right) = - k\mathop \smallint \limits_1^3 \frac{t}{{\sqrt {8 + {t^2}} }}{\rm{d}}t$ Write $u = 8 + {t^2}$. So, $du = 2tdt$. By changing the variable from $t$ to $u$, the integral becomes $V\left( P \right) = - \frac{k}{2}\mathop \smallint \limits_9^{17} {u^{ - 1/2}}{\rm{d}}u$ $V\left( P \right) = - \frac{k}{2}\cdot2\left( {{u^{1/2}}|_9^{17}} \right) = - k\left( {\sqrt {17} - 3} \right)$ With the charge density in units of ${10^{ - 6}}$ and $k = 8.99 \times {10^9}$, we obtain $V\left( P \right) = - 8.99 \times {10^9}\left( {\sqrt {17} - 3} \right) \times {10^{ - 6}} = - 10096.7$ So, the electric potential at $P$ is $V\left( P \right) = - 10096.7$ volts.
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