Question

The force exerted by an electric charge at the origin on a charged particle at a point (x, y, z) with position vector r = x, y, z is F(r) = Kr/|r|3 where K is a constant. Find the work done as the particle moves along a straight line from (4, 0, 0) to (4, 3, 4).

Answer 1

Answer:

$\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}]$

Step-by-step explanation:

The straight line path between point (a,b,c) and (l,m,n) is parametric by the expression

$r(t)=(1-t)(a,b,c) +t(l,m,n)$

since we are giving point (4,0,0) and (4,3,4), the parametric equation is giving below

$r(t)=(1-t)(4,0,0) +t(4,3,4)$

using the dot product system of multiplication, we have

$r(t)=(4,3t,4t)$

t is between 0,1.

Next we define the line integral for work done which is express as

$\int\limits^a_b {F.} \,dr$

First we define the general expression for the force

$f(x,y,z)=\frac{K(x,y,z)}{(x^{2}+y^{2}+z^{2})^{3/2}}$

If we substitute our parametric equation we arrive at

$F(r(t))=\frac{K(4,3t,4t)}{(4^{2}+(3t)^{2}+(4t)^{2})^{3/2}}\\F(r(t))=\frac{K(4,3t,4t)}{(16+34t^{2})^{3/2}}$

also we need to find the expression for dr

$r(t)=(4,3t,4t)\\dr=(0,3,4)dt$

Now we substitute into the integral expression

$\int\limits^1_0 {\frac{k(4,3t,4t)}{(16+34t^{2})^{3/2}}} \, .(0,3,4)dt$

using dot product we arrive at

$\int\limits^1_0 {\frac{25kt}{{(16+34t^{2})^{3/2}} } \,$

let make a simple substitution so we can simplify the integral,

let assume $u=16+34t^{2}$

$\frac{du}{dt}=68t\\ dt=\frac{du}{68t}$

and changing setting the new upper and lower limit, we have

$\frac{25k}{68} \int\limits^a_b {\frac{1}{u\frac{3}{2} } } \, du\\a=50\\b=16$

by simple integral we arrive at

$-\frac{25k}{34}[\frac{1}{u^{1/2}}] ^{50}_{16} \\\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}]$

Hence the workdone is

$\frac{25k}{34}[\frac{1}{4}-\frac{1}{50^{1/2}}]$