## Calculus: Early Transcendentals (2nd Edition)

$W = 2 \times {10^{ - 16}}$
$\begin{gathered} Given \hfill \\ \hfill \\ F\,\left( r \right) = \frac{{{k_q}Q}}{{{r^2}}} \hfill \\ \hfill \\ and\,\,the\,\,constats \hfill \\ q = 1.6 \times {10^{ - 19}}\,\,,\,\,\,k = 9 \times {10^9}\frac{{N - {m^2}}}{{{C^2}}} \hfill \\ \hfill \\ Q = 50q\,\,\,,\,\,\,r = 6 \times {10^{ - 11}}m \hfill \\ \hfill \\ Work\,\,\,required \hfill \\ \hfill \\ W = \int_{6 \times {{10}^{ - 11}}}^\infty {F\,\left( r \right)} \,\,dr \hfill \\ \hfill \\ substitute\,\,the\,\,values \hfill \\ \hfill \\ W = \,\left( {9 \times {{10}^9} \times 50 \times {{1.6}^2} \times {{10}^{ - 38}}} \right)\int_{6 \times {{10}^{ - 11}}}^\infty {\frac{{dr}}{{{r^2}}}} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ W = \left( {9 \times {{10}^9} \times 50 \times {{1.6}^2} \times {{10}^{ - 38}}} \right)\mathop {\lim }\limits_{b \to \infty } \,\,\left[ { - \frac{1}{r}} \right]_{6 \times {{10}^{ - 11}}}^b \hfill \\ \hfill \\ evaluate\,\,the\,\,limit \hfill \\ \hfill \\ W = \frac{{\,\left( {9 \times {{10}^9} \times 50 \times {{1.6}^2} \times {{10}^{ - 38}}} \right)}}{{6 \times {{10}^{ - 11}}}} \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ W = 2 \times {10^{ - 16}} \hfill \\ \hfill \\ \end{gathered}$