Question

The force exerted by an electric charge at the origin on a charged particle at a point \(\left( \text{x,y,z} \right)\]with a position vector\[\text{r=}\langle \text{x,y,z}\rangle \] is\[\text{F}\left( \text{r} \right)\text{=}{}^{\text{Kr}}/{}_{{{\left| \text{r} \right|}^{\text{3}}}}\] where is a constant. (See Example\[\text{5}\] in Section\[\text{13}\text{.1}\]) Find the work done as the particle moves along a straight line from\[\text{(2,0,0)to(2,1,5)}\).

Short answer

The work done is\({\rm{K}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ - }}\frac{{\rm{1}}}{{\sqrt {{\rm{30}}} }}} \right)\).

Step-by-step solution

Given.

As a result, the parametric representation of the line segment between \({\rm{(2,0,0)}}\)and \({\rm{(2,1,5)}}\)is \({\rm{r(t) = (1 - t)}} \cdot \langle {\rm{2,0,0}}\rangle {\rm{ + t}} \cdot \langle {\rm{2,1,5}}\rangle \;\)where \({\rm{t}} \in {\rm{(0,1)r(t) = }}\langle {\rm{2(1 - t),0,0}}\rangle {\rm{ + }}\langle {\rm{2t,t,5t}}\rangle \)where \({\rm{t}} \in {\rm{(0,1)r(t) = }}\langle {\rm{2,t,5t}}\rangle \;\;\)Where \({\rm{t}} \in {\rm{(0,1)}}\)

Given:

\(\begin{aligned}&{\rm{F(x,y,z) = }}\frac{{{\rm{Kr}}}}{{{\rm{|r}}{{\rm{|}}^{\rm{3}}}}}\\&{\rm{ = }}\frac{{\rm{K}}}{{{{\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}}}\langle {\rm{x,y,z}}\rangle \end{aligned}\)

Therefore,

\({\rm{F(r(t)) = }}\frac{{\rm{K}}}{{{{\left( {{{\rm{2}}^{\rm{2}}}{\rm{ + }}{{\rm{t}}^{\rm{2}}}{\rm{ + (5t}}{{\rm{)}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}}}\langle {\rm{2,t,5t}}\rangle \)

\({\rm{F(r(t)) = }}\frac{{\rm{K}}}{{{{\left( {{\rm{4 + 26}}{{\rm{t}}^2}} \right)}^{{\rm{3/2}}}}}}\langle {\rm{2,t,5t}}\rangle \)

To find work done.

Note \({\rm{dr = }}\langle {\rm{0,1,5}}\rangle {\rm{dt}}\)

Work done \({\rm{ = }}\int_{\rm{C}} {\rm{F}} \cdot {\rm{dr}}\)

Integrate from \({\rm{0}}\)to\({\rm{1}}\), because \({\rm{t}} \in {\rm{(0,1)}}\)is true.

\(\begin{aligned}\int_0^1 {\frac{{\rm{K}}}{{{{\left( {{\rm{4 + 26}}{{\rm{t}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}}}} \langle {\rm{2,t,5t}}\rangle \cdot \langle {\rm{0,1,5}}\rangle {\rm{dt = }}\int_{\rm{0}}^{\rm{1}} {\frac{{{\rm{K(t + 25t)}}}}{{{{\left( {{\rm{4 + 26}}{{\rm{t}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}}}} {\rm{dt}}\\{\rm{ = }}\frac{{\rm{1}}}{{\rm{2}}}\int_{\rm{0}}^{\rm{1}} {\frac{{{\rm{K(52t)}}}}{{{{\left( {{\rm{4 + 26}}{{\rm{t}}^{\rm{2}}}} \right)}^{{\rm{3/2}}}}}}} {\rm{dt}}\end{aligned}\)

Substitute \({\rm{4 + 26}}{{\rm{t}}^{\rm{2}}}{\rm{ = u and 52tdt = du}}\)

Integration's bounds shift over time \(\int_0^1 {{\rm{ to }}} \mathop \smallint \nolimits_4^{30} \)

\(\begin{array}{c}{\rm{ = }}\frac{{\rm{K}}}{{\rm{2}}}\int_{\rm{4}}^{{\rm{30}}} {\frac{{{\rm{du}}}}{{{{\rm{u}}^{{\rm{3/2}}}}}}} \\{\rm{ = }}\frac{{\rm{K}}}{{\rm{2}}} \cdot \left( {{\rm{ - }}\frac{{\rm{2}}}{{{{\rm{u}}^{{\rm{1/2}}}}}}} \right)_{\rm{4}}^{{\rm{30}}}\\{\rm{ = K}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ - }}\frac{{\rm{1}}}{{\sqrt {{\rm{30}}} }}} \right)\end{array}\)

Therefore, the work done is\({\rm{K}}\left( {\frac{{\rm{1}}}{{\rm{2}}}{\rm{ - }}\frac{{\rm{1}}}{{\sqrt {{\rm{30}}} }}} \right)\).